Course Content

Work Energy and Power  Lesson

Work, Energy and Power  Quiz 1 and 2
Work, Energy and Power
Introduction
The terms ‘work’, ‘energy and ‘power’ are used frequently in everyday life. A student studying for examination, mother cooking in the kitchen, a mechanic doing repair of a washing machine, are all said to be working. In Physics, ‘work’ covers a definite and precise meaning. The duration for which somebody works is called capacity. If a person puts in 1314 hours a day is said to be having high stamina. Therefore, ‘Energy’ is capacity to do work. The word ‘power’ is used in everyday life with different shades of meaning. In boxing we talk of ‘powerful’ punches. These are delivered at a great speed. This shade of meaning is close to the meaning of ‘power’ used in physics. Let us find out the loose correlation between the physical definitions and physiological pictures these terms generate in our minds. This chapter encapsulates the understanding of these physical quantities. Prior to that, we have to develop a mathematical prerequisite, namely the scalar product of two vectors.
The Scalar Product
We have learnt about vectors and their use in earlier lesson ‘Motion in a Plane’ and we know that physical quantities like displacement, velocity, acceleration, force etc., are vectors and how they are added or subtracted. Let us find out how vectors are multiplied. There are two ways of multiplying vectors:
 The scalar product gives a scalar from two vectors.
 Vector product produces a new vector from two vectors.
We shall learn about vector product later on. The scalar product or dot product of any two vectors A and B denoted as A.B (read as A dot B) is defined as
A.B = AB cos θ ………………………………………………………………………………………………(1a)
where θ is the angle between the two vectors as shown in Fig. 1a. Since A, B and cos θ are scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction but their scalar product does not have a direction.
From Eq. (1a), we have
A.B = A(B cos θ)
= B(A cos θ)
Geometrically, B cos θ is the projection of B onto A Fig. 1(b) and A cos θ is the projection of A onto B in Fig. 1(c). So, A.B is the product of the magnitude of A and the component of B along A. Alternatively, it is the product of the magnitude of B and the component of A along B.
Fig. 1 (a) The scalar product of two vectors A and B is a scalar: A.B = A B cos θ (b) B cos θ is the projection of B on to A (c) A cos θ is the projection of A on to B.
Equation (1a) shows that the scalar product follows the commutative law:
A.B = B.A
Scalar product obeys the distributive law:
A.(B+C) = A.B + A.C
Further, A.(λB) = λ (A.B) where λ is a real number.
The proofs of the above equations are left to you as an exercise.
For unit vectors i, j, k we have,
i.i = j.j = k.k = 1
i.j = j.k = k.i = 0
Given two vectors,
A = A_{x} i + A_{y}j + A_{z}k
B = B_{x} i + B_{y}j + B_{z}k
their scalar product is
A.B = (A_{x} i + A_{y}j + A_{z}k).(B_{x} i + B_{y}j + B_{z}k)
= A_{x} B_{x} + A_{y}B_{y} + A_{z}B_{z ………………………………………………………………………………………………}_{(1b)}
From the definition of scalar product and Eq. 1(b) we have:
(i) A.A = A_{x}A_{x} + A_{y}A_{y} + A_{z}A_{z}
Or, A^{2 }= (A_{x})^{2} + (A_{y})^{2} + (A_{z})^{2}_{ ……………………………………………………………………………………}_{(1c)}
since A.A = I A I A I cos 0 = A^{2}
(ii) A.B = 0, if A and B are perpendicular.
Find the angle between force F = (3i +4j – 5k) unit and displacement d = (5i + 4j +3k) unit. Also find the projection of F and d.
Answer
F.d = F_{x}d_{x} + F_{y}d_{y} + F_{z}d_{z}
= 3(5) + 4(4) + (5)(3)
= 16
Hence F.d =F d cos θ = 16 unit
Now F.F = F^{2 }= (F_{x})^{2} + (F_{y})^{2} + (F_{z})^{2}
= 9 +16 +25
= 50 unit
and d.d = d^{2 }= (d_{x})^{2} + (d_{y})^{2} + (d_{z})^{2}
= 25 +16 +9
= 50 unit
Therefore, cos θ = 16/(√50 √50) = 16/50 = 0.32
θ = cos^{1}0.32
Notions of Work and Kinetic Energy: The WorkEnergy Theorem
The following relation for rectilinear motion under constant acceleration a has been encountered in previous lesson.
v^{2 }– u^{2} = 2as where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have
1/2 mv^{2 }1/2 mu^{2 }= mas = Fs …………………………………………………………………….(2a)
where the last step follows from Newton’s Second Law. We can generalise Eq. 1 to three dimensions by employing vectors
v^{2 }– u^{2} = 2 a.d
Once again multiplying both sides by m/2, we obtain
1/2 mv^{2 }1/2 mu^{2 }= ma.d = F.d ………………………………………………………………..(2b)
The above equation provides a motivation for definitions of work and kinetic energy. The left side of the equation is the difference in the quantity ‘half the mass times the square of the speed’ from its initial value to its final value. We call each of these quantities the ‘kinetic energy’, denoted by K . The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by W. Eq. (2) is then
K_{f }– K_{i }= W ……………………………………………………………………………………….(3)
where K_{f }and K_{i }are respectively the final and initial kinetic energies of the object. Work refers to the force and displacement over which it acts. Work is done by a force on the body over a displacement.
Equation (2) is also a special case of the workenergy (WE) theorem: The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to varying force in a later section.
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.oo km. It hits the ground with a speed of 50.0 m s^{–}^{1}. What is the work done by the unknown resistive force?
Answer
(a) The change in kinetic energy of the drop is
ΔK = 1/2 mv^{2 }–0
= 1/2 × 10^{–}^{3}× 50 × 50
= 1.25 J where we have assumed that the drop is initially at rest.
Assuming that g is a constant with a value 10 m/s^{2 }, the work done by the gravitational force is,
W_{g }= mgh
= 10^{–}^{3}× 10 × 10^{3}
= 10.0 J
(b) From the workenergy theorem
ΔK = W_{g }+ W_{r }where W_{r }is the work done by the resistive force on the rain drop. Thus
W_{r } = ΔK – W_{g }
= 1.25 – 10
=8.75 J is negative
Work
As seen earlier, work is related to force and the displacement over which it acts. Consider a constant force F acting on an object of mass m. The object undergoes a displacement d in the positive xdirection as shown in Fig.2.
The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus
W = (F cos θ)d = F.d …………………………………………………………………………………(4)
We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, the force you exert on the wall does no work. Yet your muscles are alternatively contracting and relaxing and internal energy is being used up and you do get tired. Thus, the meaning of work in physics is different from its usage in everyday language.
No work done if:
(i) the displacement is zero as seen in the example above. A weight lifter holding a 150 kg mass steadily on his shoulder for 30 s does no work on the load during this time.
(ii) the force is zero. A block moving on a smooth horizontal table is not acted upon by a horizontal force (since there is no friction), but may undergo a large displacement.
(iii) the force and displacement are mutually perpendicular. This is so since θ = π/2 rad (= 90^{0}), cos (π/2) = 0. For the block moving on a smooth horizontal table, the gravitational force mg does no work since it acts at right angles to the displacement. If we assume that the moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and θ = π/2.
Work can be both positive and negative. If θ is between 0^{0 }and 90^{0 }, cos θ in Eq. (4) is positive. If θ is between 90^{0 }and 180^{0 }, cos θ in Eq. (4) is negative. In many examples the frictional force opposes displacement and θ = 180^{0}. Then the work done by friction is negative (cos 180^{0}= 1).
From Eq. (4) it is clear that work and energy have the same dimensions, [ML^{2}T^{2}]. The SI unit of these is joule (J). named after the famous British physicist James Prescott Joule (181111869). Since work and energy are so widely used as physical concepts, alternative units abound and some of these are listed in Table1 below:
Table1 Alternative Units of Work/Energy in J
Name of Unit  Conversion to Joule 

erg  multiply by 10^{7} 
electron volt (eV)  multiply by 1.6 × 10^{19} 
calorie (cal)  multiply by 4.186 
kilowatt hour (kWh)  multiply by 3.6×10^{6} 
Example 3
A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Answer
Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
(a) The stopping force and the displacement make an angle of 180^{0 }(π rad) with each other.
Thus, work done by the road,
W_{r} = Fd cos θ
= 200 × 10 × cos π
= 2000J
It is this negative work that brings the cycle to halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus work done by cycle on the road is zero.
The lesson of this example is that though the force on a body A exerted by the body B is always equal and opposite to that on B by A (Newton’s Third Law); the work done on A by B is not necessarily equal and opposite to the work done on B by A.
Kinetic Energy
As noted earlier, if an object of mass m has velocity v, its kinetic energy K is
K = 1/2 m v.v = 1/2mv^{2} ……………………………………………………………………(5)
Kinetic Energy is a scalar quantity. The kinetic energy of an object is a measure of the work an object can do by the virtue of its motion. This notion has been intuitively known for a long time. The kinetic energy of a fast flowing stream has been used to grind corn. Sailing ships employ the kinetic energy of the wind. Table 2 lists the kinetic energies for various objects.
Table2 Typical kinetic energies (K)
Object  Mass (kg)  speed (ms^{1})  K(J) 

Car  200  25  6.3 × 10^{5} 
Running athlete  70  10  3.5 × 10^{3} 
Bullet  5×10^{2}  200  10^{3} 
Stone dropped from 10 m  1  14  10^{2} 
Rain drop at terminal speed  3.5 × 10^{5}  9  1.4×10^{3} 
Air molecule  ≅ 10^{26}  500  ≅ 10^{21} 
Example 4
In a ballistics demonstration a police officer fires bullet of mass 50.0 g with speed 200 m s^{1} (See Table 2) on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet?
Answer
The initial kinetic energy 1/2 mv^{2 }= 1000 J. It has a final kinetic energy of 0.1 × 1000 = 100 J. If v_{f }is the emergent speed of the bullet,
1/2 m(v^{2})_{f}^{ }= 100 J
v_{f }= √(2 × 100 J)/0.05kg
= 63.2 ms^{1}
The speed is reduced by approximately 68% (not 90%)
Work done by a Variable Force
A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 2 is a plot of a varying force in one dimension.
If the displacement Δx is small, we can take the force F(x) as approximately constant and the work done is then
ΔW = F(x) Δx
This is illustrated in Fig. 3(a). Adding successive rectangular areas in Fig. 3(a), we get the total work done as,
W =∑^{xf}_{xi} F(x)Δx ……………………………………………………………………………(6)
where the summation is from the initial position x_{i }to the final position x_{f}
If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 3(b). Then the work done is
W = lim_{Δx→0}∑^{xf}_{xi} F(x)Δx
= ∫^{xf}_{xi} F(x)dx ……………………………………………………………………………….(7)
where ‘lim’ stands for the limit of the sum when Δx tends to zero. Thus, for a varying force the work done can be expressed as a definite integral of force over displacement
Fig. 3 (a) The shaded rectangle represents the work done by the varying force F(x), over the small displacement Δx, ΔW = F(x)Δx. (b) adding the areas of all the rectangles we find that Δx→0, the area under the curve is exactly equal to the work done by F(x).
Example 5
A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 5O N. The total distance through which the trunk has been removed is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N. Calculate the work done by the two forces over 20 m.
Answer
Fig. 4 Plot of the force F applied by the woman and the opposing frictional force f.
The plot of the applied force is shown in Fig. 4 above. At x = 20 m, F = 50 N(≠0). We are given that the frictional force f is I f I = 50 N. It opposes motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis.
The work done by the woman is
W_{F} →area of the rectangle ABCD + area of the trapezium CEID
W_{F} = 100 × 10 + 1/2(100+50) ×10
= 1000 + 750
= 1750 J
The work done by the frictional force is
W_{f} →area of the rectangle AGHI
W_{f} = (50) × 20
=1000 J
The area on the negative side of the force axis has negative sign.
The WorkEnergy Theorem for a Variable Force
We are now familiar with the concepts of work and kinetic energy to prove the workenergy theorem for a variable force. We confine ourselves to one dimension. The time rate of change of kinetic energy is
dK/dt = d/dt (1/2mv^{2})
= m (dv/dt)v
= F v (from Newton’s Second Law)
= F dx/dt
Thus
dK = Fdx
Integrating from the initial position (x_{i}) to final position (x_{f}), we have
∫^{Kf}_{Ki}dK = ∫^{xf}_{xi}Fdx where K_{i} and K_{f }are the initial and final kinetic energies corresponding to x_{i} and x_{f }
or K_{f }–K_{i} = ∫^{xf}_{xi}Fdx …………………………………………………………………….(8a)
From Eq. (7), it follows that,
K_{f }–K_{i} = W ………………………………………………………………………………..(8b)
Thus, the WE theorem is useful in a variety of problems, it does not, in general, incorporate the complete dynamical information of Newton’s second law. It is an integral form of Newton’s second law. Newton’s second law is a relation between acceleration and force at any instant of time. In this sense, the temporal (time) information contained in the statement of Newton’s second law is ‘integrated over’ and is not available explicitly. Another observation is that Newton’s second law for two or three dimensions is in vector form. In the scalar form, information with respect to directions contained in Newton’s second law is no present.
Example 6
A block of mass m = 1 kg, moving on a horizontal surface with speed v_{i} = 2 ms^{1} enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force F_{r} on the block in this range is inversely proportional to x over this range,
Fr = k/x for 0.1 < x < 2.01 m
= 0 for x < 0.1 m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed v_{f} of the block as it crosses the pitch.
Answer
From Eq. (8a)
Here, note that ln is a symbol for the natural logarithm to the base e and not the logarithm to the base 10[ln X = log_{e}X = 2.303 log_{10}X].
The Concept of Potential Energy
The word potential suggests possibility or capacity for action. The term potential energy brings to one’s mind ‘stored’ energy. A stretched bowstring possesses potential energy. When it is released, the arrow flies off at a great speed. The earth’s crust is not uniform, but has discontinuities and dislocations that are called fault lines. These fault lines in the earth’s crust are like ‘compressed springs’. They possess a large amount of potential energy. An earthquake results when these fault lines readjust. Thus, potential energy is the ‘stored energy’ by virtue of the position or configuration of a body. The body left to itself releases this stored energy in the form of kinetic energy. Let us make our notion of potential energy more concrete.
The gravitational force on a ball of mass m is mg, g may be treated as a constant near the earth surface. By ‘near’ we imply that the height h of the ball above earth’s surface is very small compared to the earth’s radius R_{2} (h«R_{2}) so that we can ignore the variation of g near the earth’s surface. In what follows we have taken the upward direction to be positive. Let us raise the ball up to a height h. The work done by the external agency against the gravitational force is mgh. This work gets stored as potential energy. Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it is the negative of work done by the gravitational force in raising the object to that height.
V(h) = mgh
If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of the derivative V(h) with respect to h. Thus,
F = d/dh(V(h) = mg
The negative sign indicates that the gravitational force is downward. When released, the ball comes down with an increasing speed. Just before it hits the ground, its speed is given by the kinematic relation,
v^{2}= 2gh
This equation can be written as
1/2 mv^{2}= mgh which shows that the gravitational potential energy of the object at height h, when the object is released, manifests itself as kinetic energy of the object on reaching the ground.
Physically, the notion of potential energy is applicable only to the class of forces where work done against the force gets ‘stored up’ as energy. When external constraints are removed, it manifests itself as kinetic energy. Mathematically, (for simplicity, in one dimension) the potential energy V(x) is defined if the force F(x) can be written as
F(x) = –dV/dx
This implies that
∫^{xf}_{xi}F(x)dx = ∫^{vf}_{vi}dV = V_{i }– V_{f}
The work done by a conservative force such as gravity depends on the initial and final positions only. In the previous chapter we have worked on examples dealing with inclined planes. If an object of mass m is released from rest, from the top of a smooth (frictionless) inclined plane of height h, its speed at the bottom is √2gh irrespective of the angle of inclination. Thus, at the bottom of the inclined plane it acquires a kinetic energy, mgh. If the work done or the kinetic energy did depend on other factors such as the velocity or the particular path taken by the object, the force would be called nonconservative.
The dimensions of potential energy are [ML^{2}T^{2}] and the unit is joule (J). the same as kinetic energy or work. To reiterate, the change in potential energy, for a conservative force,
ΔV = F(x)Δx ……………………………………………………………………………..(9)
In the example of the falling ball considered in this section we saw how potential energy was converted to kinetic energy. This hints at an important principle of conservation in mechanics, which we now proceed to examine.
The Conservation of Mechanical Energy
For simplicity we demonstrate this important principle for one dimensional motion. Suppose that a body undergoes displacement Δx under the action of a conservative force F. Then from the WE theorem we have,
ΔK = F(x)Δx
If the force is conservative, the potential energy function V(x) can be defined such that
–ΔV = F(x)Δx
The above equations imply that
ΔK + ΔV = 0
Δ(K + V) = 0 ………………………………………………………………………………..(10)
which means that K + V, the sum of the kinetic and potential energies of the body is constant. Over the whole path, x_{i }to x_{f}, this means that
K_{i} + V(x_{i}) = K_{f} + V(x_{f}) ………………………………………………………………….(11)
The quantity K + V(x), is called the total mechanical energy of the system. Individually the kinetic energy K and the potential energy V(x) may vary from point to point, but the sum is a constant. The aptness of the term ‘conservative force’ is now clear.
Let us consider some of the definitions of a conservative force.
 A force F(x) is a conservative if it can be derived from a scalar quantity V(x) by the relation given by Eq. (9). The three dimensional generalization requires the use of a vector derivative, which is outside the scope of this book.
 The work done by the conservative force depends only on the end points. This can be seen from the relation, W = K_{f }– K_{i }= V(x_{i }) – V(x_{f }) which depends on the end points.
 A third definition states that the work done by this force in a closed path is zero. This is once again apparent from Eq. (11) since x_{i }=x_{f .}
Thus, the principle of conservation of total mechanical energy can be stated as
The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.
The above discussion can be made more concrete by considering the example of the gravitational force once again and that of the spring force in the next section. Fig. 5 depicts a ball of mass m being dropped from a cliff of a height H.
Fig. 5 The conversion of potential energy to kinetic energy for a ball of mass m dropped from a height H
The total mechanical energies E_{0}, E_{h} and E_{H} of the ball at the indicated height zero (ground level), h and H are
E_{H} = mgH …………………………………………………………………………………….(11a)
E_{h} = mgh + 1/2 m(v_{h})^{2}…………………………………………………………………….(11b)
E_{0} = 1/2 mv^{2}……………………………………………………………………………………(11c)
The constant force is a special case of spatially dependent force F(x). Hence, the mechanical energy is conserved. Thus
E_{H} =E_{0}
or, mgH = 1/2 m(v_{f})^{2}
v_{f }= √2gH a result that was obtained for a free falling body.
Further,
E_{H} =E_{h }
which implies,
(v_{h})^{2}=2g(Hh) ………………………………………………………………………………..(11d)
and is familiar result from kinematics.
At the height H, the energy is purely potential. It is partially converted to kinetic at height h and is fully kinetic at ground level. This illustrates the conservation of mechanical energy.
Example 7
A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_{o} at the lowest point A such that it completes a semicircular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point C. This is shown in Fig.6. Obtain an expression for (i) v_{o} ; (ii) the speeds at points B and C; (iii) the ration of kinetic energies K_{B}/K_{C} at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.
Answer
(i) There are two external forces on the bob : gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A:
E = 1/2(mv_{0})^{2} …………………………………………………………………………………………(12)
T_{A }– mg = m(v_{0})^{2}/L [Newton’s Second Law]
where T_{A }is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string (T_{A}) becomes zero.
Thus, at C
E= 1/2m(v_{c})^{2} + 2mgL …………………………………………………………………………..(13)
mg = m(v_{c})^{2}/L [Newton’s Second Law] ……………………………………………………(14)
where v_{c }is the speed at C. From Eqs. (13) and (14)
E = 5/2 mgL
Equating this energy at A
5/2 mgL = 1/2 m(v_{0})^{2}
v_{0}= √5gL
(ii) It is clear from Eq. (14)
v_{c}= √gL
At B, the energy is
E = 1/2m(v_{B})^{2} + mgL
Equating this to the energy at A and employing the result from (i), namely v_{0}^{2} = 5gL,
1/2m(v_{B})^{2} + mgL = 1/2m(v_{0})^{2}
=(5/2)mgL
∴ v_{B}= √3gL
(iii) The ratio of the kinetic energies at B and C is:
K_{B}/K_{C} = 1/2m(v_{B})^{2}/1/2m(v_{C})^{2}=3/1
At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete revolution.
Potential Energy of a Spring
The spring force is an example of a variable force which is conservative. Fig. 7 shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force F_{s }is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 7(b)] or negative [Fig. 7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as
F_{s }= kx
The constant k is called the spring constant. Its unit is Nm^{1}. The spring is said to be stiff if k is small.
Suppose that we pull the block outwards as in Fig. 7(b). If the extension is x_{m}, the work done by the spring force is
W_{s} = ∫_{0}^{xm}F_{5}dx =∫_{0}^{xm}kxdx
= –k(x_{m})^{2}/2………………………………………………………………………………………………(15)
This expression may also be obtained by considering the area of the triangle as in Fig. 7(d). Note that the work done by the external pulling force F is positive since it overcomes the spring force.
W = +k(x_{m})^{2}/2………………………………………………………………………………………………(16)
Fig. 7 Illustration of the spring force with a block attached to the free end of the spring.
(a) The spring force F_{s} is zero when the displacement x from the equilibrium position is zero.
(b) For the stretched spring x > 0 and F_{s}< 0
(c) For the compressed spring x < 0 and F_{s}> 0
(d) The plot of F_{s }versus x.
The area of the shaded triangle represents work done by the spring force. Due to the opposing signs of F_{s }and x this work done is negative, W_{s} = k(x_{m})^{2}/2.
The same is true when the spring is compressed with a displacement x_{c}(<0). The spring force does work W_{s} = –k(x_{c})^{2}/2 while the external force F does the work +k(x_{c})^{2}/2. If the block is moved from an initial displacement x_{i }to a final displacement x_{f }, work done by the spring force W_{s }= ∫_{xi}^{xf}kxdx = k (x_{i})^{2}/2 – k (x_{f})^{2}/2 ………………………………..(17)
Thus the work done by the spring force depends only on the endpoints. Specifically, if the block is pulled from x_{i }and allowed to return to x_{i }:
W_{s }= ∫_{xi}^{xi}kxdx = k (x_{i})^{2}/2 – k (x_{i})^{2}/2 ………………………………..(18)
=0
The work done by the spring force in a cyclic process is zero. We have explicitly demonstrated that the spring force (i) is position dependent only as first stated by Hooke. (F_{s }= kx): (ii) does work which only depends on the initial and final positions, e.g. Eq. (17). Thus, the spring force is a conservative force.
We define the potential energy V(x) of the spring to be zero when block and spring system is in the equilibrium position. For an extension (or compression) x the above analysis suggests that
V(x) = kx^{2}/2 ……………………………………………………………………….(19)
You may easily verify that dV/dx = –kx, the spring force. If the block of mass m in Fig. 7 is extended to x_{i }and released from rest, then its total mechanical energy at any arbitrary point x, where x lies between –x_{m }and +x_{m }will be given by
1/2k(x_{m})^{2}= 1/2kx^{2 }+ 1/2mv^{2}
where we have invoked the conservation of mechanical energy. This suggests that the speed and kinetic energy will be maximum at the equilibrium position, x = 0. i.e.,
1/2m(v_{m})^{2}= 1/2k(x_{m})^{2}
where v_{m }is the maximum speed.
or v_{m }=[ √(k/m)]x_{m}
Note that k/m has the dimensions of [ T]^{2} and our equation is dimensionally correct. The kinetic energy converted to potential energy and vice versa, however, the total mechanical energy remains constant. This is graphically depicted in Fig. 8.
Example 8
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10³ C. What is the maximum compression of the spring.
Fig. 8 Parabolic plots of the potential energy V and kinetic energy K of a block attached to a spring obeying Hooke’s law.
The two plots are complementary, one decreasing as the other increases. The total mechanical energy E = K +V remains constant.
Answer
At maximum compression the kinetic energy of the car is converted entirely in to the potential energy of the spring.
The kinetic energy of the moving car is
K = 1/2mv^{2}
= 1/2 × 10³ × 5 × 5
K = 1.25 × 10^{4 }J
where we have converted 18 km h^{1 }to 5 m s^{1} [It is useful to remember that 36 km h^{1 }= 10 m s^{1} ]. At maximum compression x_{m}, the potential energy V of the spring is equal to the kinetic energy K of the moving car from the principle of conservation of mechanical energy,
V = 1/2k(x_{m})^{2}
= 1.25 × 10^{4 }J
We obtain
x_{m }= 2.00 m
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.
We conclude this section by making a few remarks on conservative forces.
 Information on time is absent from the above discussions. In the example considered above, we can calculate the compression, but not the time over which the compression occurs. A solution of Newton’s Second Law for this system is required for temporal information.
 Not all forces are conservative. Friction, for example, is a nonconservative force. The principle of conservation of energy will have to be modified in this case. This is illustrated in Example 9.
 The zero of the potential energy is arbitrary, It is set according to convenience. For the spring force we took V(x) = 0, at x = 0. i.e., the unstretched spring had zero potential energy. For the constant gravitational force mg, we took V = 0 on the earth’s surface. In a later lesson we shall see that the force due to the universal law of gravitation, the zero is the best defined at an infinite distance from the gravitational source. However, once the zero of the potential energy is fixed in a given discussion, it must be consistently adhered to throughout the discussion. You cannot change the horses in the midstream!
Consider Example 7 taking the coefficient of friction, μ, to be 0.5 and calculate the maximum compression of the spring.
Answer
In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Fig. 9. We invoke the workenergy theorem, rather than the conservation of mechanical energy.
The change in kinetic energy is
ΔK = K_{f} – K_{i} = 0 – 1/2 mv^{2}
The work done by the net force is
W = 1/2 k(x_{m})^{2} – μmgx_{m}
Equating we have,
1/2 mv^{2} = 1/2 k(x_{m})^{2} + μmgx_{m}
Now μmg = 0.5 × 10³ × 10 = 5 × 10³ N (taking g = 10.0 m s^{2} ). After rearranging the above equation, we obtain the following quadratic equation in the unknown x_{m}.
k(x_{m})^{2} +2μmgx_{m }– mv^{2} = 0
x_{m }= {μmg + [μ^{2}m^{2}g^{2} + mkv^{2}]^{1/2}}/k where we take the positive square root since x_{m }is positive. Putting in numerical values we obtain
x_{m }= 1.35 m
which, as expected, is less than the result in Example 8.
If the two forces on the body consist of a conservative force F_{c }and a nonconservative force F_{nc }, the conservation of mechanical energy formula will have to be modified. By the WE theorem,
(F_{c }+ F_{nc })Δx = ΔK
But F_{c }Δx = ΔV
Hence, Δ(K+V) = F_{nc }Δx
ΔE = F_{nc }Δx
where E is the total mechanical energy. Over the path this assumes the form
E_{f }– E_{i }= W_{nc}
where W_{nc} is the total work done by the nonconservative forces over the path. Note that unlike the conservative force, W_{nc }depends on the partiular path i to f.
Various forms of Energy: The Law of Conservation of Energy
In the previous section we have discussed mechanical energy. We have seen that it can be classified into two distinct categories:
 one based on motion, namely kinetic energy;
 the other on configuration (position), namely potential energy.
Energy comes in many a forms which transform into one another in ways which may not often be clear to us.
Heat
We have seen that the frictional force is excluded from the category of conservative forces. However, work is associated with the force of friction. A block of mass m sliding on a rough horizontal surface with speed v_{0 }comes to halt over a distance x_{0}. The work done by the force of kinetic friction f over x_{0 }is –fx_{0}. By the workenergy theorem 1/2 m(v_{0})^{2}=f x_{0}. If we confine our scope to mechanics, we would say that the kinetic energy of the block is ‘lost’ due to the frictional force. One examination of the block and the table we would detect a slight increase in their temperatures. The work done by friction is not ‘lost’, but is transferred as heat energy. This raises the internal energy of the block and the table. In winter, in order to feel warm, we generate heat by vigorously rubbing our palms together. We shall see later that the internal energy is associated with the ceaseless, often random, motion to molecules. A quantitative idea of the transfer of heat energy is obtained by noting that 1 kg of water releases about 42000 J of energy when it cools by 10º C.
Chemical Energy
One of the greatest technical achievements of humankind occurred when we discovered how to ignite and control fire. We learnt to rub two flint stones together (mechanical energy), got them to heat up and to ignite a heap of dry leaves (chemical energy), which then provided sustained warmth. A matchstick ignites in to a bright flame when struck against a specially prepared chemical surface. The lighted matchstick, when applied to a fire cracker, results in a spectacular display of sound and light.
Chemical energy arises from the fact that the molecules participating in the chemical reaction have different binding energies. A stable chemical compound has less energy than the separated parts. A chemical reaction is basically a rearrangement of atoms. If the total energy of the reactants is more than the products of the reaction, heat is released and the reaction is said to be exothermic reaction. If the reverse is true, heat is absorbed and the reaction is endothermic. Coal consists of carbon and a kilogram of it when burnt releases 3 × 10^{7} J of energy.
Chemical energy is associated with the forces that give rise to the stability of substances. These forces bind atoms in to molecules, molecules in to polymeric chains, etc. The chemical energy arising from the combustion of coal, cooking gas, wood and petroleum is indispensable to our daily existence.
Electrical Energy
The flow of electrical current causes bulbs to glow, fans to rotate and bells to ring, There are laws governing the attraction and repulsion of charges and currents, which we shall cover up later. Energy is associated with an electric current. An urban Indian household consumes about 200 J of energy per second on an average.
The Equivalence of Mass and Energy
Till the end of the nineteenth century, physicists believed that in every physical and chemical process, the mass of an isolated system is conserved. Matter might change its phase, e.g., glacial ice could melt in to gushing stream, but matter is neither created nor destroyed; Albert Einstein (18791955) however, showed that mass and energy are equivalent and related by the relation
E = mc^{2} ……………………………………………………………………….(20)
where c, the speed of light in vacuum is approximately 3 × 10^{8 }m s^{1}. Thus, a staggering amount of energy is associated with a mere kilogram of matter
E = 1 ×(3 × 10^{8})^{2 }J = 9 × 10^{16 }J.
This is equivalent to the annual electrical output of a large (3000 MW) power generating station.
Nuclear Energy
The most destructive weapons made by man, the fission and fusion bombs are manifestations of the above equivalence of mass and energy (Eq. 20). On the other hand the explanation of the lifenourishing energy output of the sun is also based on the above equation. In this case effectively four light hydrogen nuclei fuse to form a helium nucleus whose mass is less than the sum of the masses of the reactants. This mass difference, called the mass effect Δm is the source of energy (Δm)c^{2}. In fission, a heavy nucleus like uranium is split by a neutron in to lighter nuclei. Once again the final mass is less than the initial mass and the mass difference translates in to energy, which can be tapped to provide electrical energy as in nuclear power plants (controlled nuclear fission) or can be employed in making nuclear weapons (uncontrolled nuclear fission). Strictly, the energy ΔE released in a chemical reaction can also be related to the mass defect Δm = ΔE/c^{2}. However , for a chemical reaction, this mass defect is much smaller than for a nuclear reaction. Table 3 lists the total energies for a variety of events and phenomena.
Table3 Approximate energy associated with various phenomena
Description  Energy (J) 

Big Bang  10^{68} 
Radio Energy emitted by the galaxy during its lifetime  10^{55} 
Rotational Energy of the Milky Way  10^{52} 
Energy released in a supernova explosion  10^{44} 
Ocean’s hydrogen in fusion  10^{34} 
Rotational energy of the earth  10^{29} 
Annual solar energy incident on the earth  5×10^{24} 
Annual wind energy dissipated near earth’s surface  10^{22} 
Annual global energy usage by human  3×10^{20} 
Annual energy dissipated by the tides  10^{20} 
Energy release of 15megaton fusion bomb  10^{17} 
Annual electrical output of large generating plant  10^{16} 
Thunderstorm  10^{15} 
Energy released in burning 1000 kg coal  3×10^{10} 
Kinetic energy of a large jet aircraft  10^{9} 
Energy released in burning 1 litre of gasoline  3×10^{7} 
Daily food intake of a human adult  10^{7} 
Work done by a human heart per beat  0.5 
Turning this page  10^{3} 
Flea hop  10^{7} 
Discharge of a single neuron  10^{10} 
Typical energy of a proton in a nucleus  10^{13} 
Typical energy of an electron in an atom  10^{18} 
Energy to break one bond in DNA  10^{20} 
Example 10
Examine Tables 1 to 3 and express (a) The energy required to break one bond in DNA in eV: (b) The kinetic energy of an air molecule (10^{21}J) in eV: (c) The daily intake of human adult in kilo calories.
Answer
(a) Energy required to break one bond of DNA is
10^{20}J/(1.6 ×10^{19}J/eV) ≅ 0.06 eV where the symbol ‘≅’ stands for approximate. Note 0.1 eV =100 meV (100 millielectron volt).
(b) The kinetic energy of an air molecule is
10^{21}J/(1.6 ×10^{19}J/eV) ≅ 0.0062 eV
This is the same as 6.2 meV.
(c) The average human consumption in a day is
10^{7}J/(4.2 × 10^{3}J/kcal) ≅ 2400 kcal
We point out a common misconception created by news papers and magazines. They mention food values in calories and urge us to restrict diet intake to below 2400 calories. What they should be saying is kilo calories (kcal) and not calories. A person consuming 2400 calories a day will soon starve to death! 1 food calorie is 1 kcal.
The Principle of Conservation of Energy
We have seen that the total mechanical energy of the system is conserved if the forces doing work on it are conservative. If some of the forces involved are nonconservative, part of the mechanical energy may get transformed into other forms such as heat, light and sound. However, the total energy of an isolated system does not change, as long as one accounts for all forms of energy. Energy may be transformed from one form to another but the total energy of an isolated system remains constant. Energy can neither be created, nor destroyed.
Since the universe as a whole may be viewed as an isolated system, the total energy of the universe is a constant. If one part of the universe loses energy, another part must gain an equal amount of energy.
The principle of conservation of energy cannot be proved. However, no violation of this principle has been observed. The concept of conservation and transformation of energy in to various forms links together various branches of physics, chemistry and life sciences. It provides a unifying, enduring element in our scientific pursuits. From engineering point of view all electronic, communication and mechanical devices rely on some forms of energy transformation.
Power
Often it is interesting to know not only the work done on an object, but also the rate at which this work is done. We say a person is physically fit if he not only climbs four floors of a building but climbs them fast. Power is defined as the time rate at which work is done or energy is transferred.
The average power of a force is defined as the ratio of work, W, to the total time t taken
P_{av} = W/t
The instantaneous power is defined as the limiting value of the average power is defined as the limiting value of the average power as time interval approaches zero.
P = dW/dt ………………………………………………………………….(21)
The work dW done by a force F for a displacement dr is dW = F.dr. The instantaneous power can also be expressed as
P =F.dr/dt
=F.v …………………………………………………………………………………..(22)
where v is the instantaneous velocity when the force is F.
Power, like work and energy, is a scalar quantity. Its dimensions are [ML^{2}T^{3}]. In the SI, its unit is called watt (W). The watt is 1 J s^{1}. The unit of power is named after James Watt, one of the innovators of the steam engine in the eighteenth century.
There is another unit of power, namely the horsepower (hp)
1 hp = 746 W
This unit is still used to describe the output of automobiles, motorbikes etc.
We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy.
100 (watt) × 10 (hour) = 1000 watt hour
= 1 kilowatt hour (kWh)
= 10^{3} (W) × 3600 (s)
=3.6 × 10^{3} J
Our electricity bills carry the energy consumption in kWh. Note that kWh is a unit of energy and not power.
Example 11
An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 m s^{1}. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer
The downward force on the elevator is F = mg + F_{f }= (1800 × 10) + 4000 = 22000 N
The motor must supply enough power to balance this force. Hence,
P = F.v = 22000 × 2 = 44000 W = 59 hp
Collisions
In physics we study motion (change in position). At the same time, we try to discover physical quantities, which do not change in a physical process. The laws of momentum and energy conservation are typical examples. In this section we shall apply these laws to a commonly encountered phenomena, namely collisions. Several games such as billiards, marbles or carrom involve collisions. We shall study the collision of two masses in an idealised form.
Consider two masses m_{1 }and m_{2}. The particle m_{1 }is moving with speed v_{1i }. The subscript i_{ }implying initial. We can consider m_{2 }to be at rest. No loss of generality is involved in making such a selection. In this situation the mass m_{1 }collides with the stationary mass m_{2 }and this is depicted in Fig. 10.
The masses m_{1 }and m_{2 }flyoff in different directions. We shall see that there are relationships, which connect the masses, the velocities and the angles.
Elastic and Inelastic Collisions
In all collisions the total linear momentum is conserved; the initial momentum of the system is equal to the final momentum of the system. One can argue this as follows. when two objects collide, the mutual impulsive forces acting over the collision time Δt cause a change in their respective momenta:
Δp_{1} = F_{12}Δt
Δp_{2} = F_{21}Δt
where F_{12 }is the force exerted on the first particle by the second particle. F_{21 }is likewise the force exerted on the second particle by the first particle. Now from Newton’s Third Law, F_{12 }= –F_{21}. This implies
Δp_{1} + Δp_{2} = 0
The above conclusion is true even though the forces vary in a complex fashion during the collision time Δt. Since the third law is true at every instant, the total impulse on the first object is equal and opposite to that on the second.
On the other hand, the total kinetic kinetic energy of the system is not necessarily conserved. The impact and deformation during collision may generate heat and sound. Part of the initial kinetic energy is not transformed in to other forms of energy. A useful way to visualize the deformation during collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy during the collision time Δt is not constant. Such a collision is called an elastic collision. On the other hand the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision.
Collisions in One Dimension
Consider first a completely inelastic collision in one dimension. Then, in Fig. 10,
θ_{1} = θ2 = 0
m_{1}v_{1i} = (m_{1} +m_{2})v_{f }(momentum conservation)
v_{f} = [m_{1} /(m_{1}+m_{2})]v_{1i} ……………………………………………………..(23)
The loss in kinetic energy on collision is
ΔK = 1/2m_{1}(v_{1i})^{2}1/2 (m_{1}+m_{2})(v_{1f})^{2}
=1/2m_{1}(v_{1i})^{2}1/2 (m_{1}^{2}/m_{1}+m_{2})(v_{1i})^{2 }using Eq. (23)
=1/2m_{1}(v_{1i})^{2}[1m_{1} /(m_{1}+m_{2})]
= 1/2 [(m_{1}m_{2})/(m_{1}+m_{2})](v_{1i})^{2}
which is a positive quantity as expected.
Consider next an elastic collision. Using the above nomenclature with θ_{1} = θ2 = 0, the momentum and kinetic energy conservation equations are
m_{1}v_{1i} = (m_{1} v_{1f}+m_{2}v_{2f }) ……………………………………………………….(24)
m_{1}(v_{1i})^{2} = (m_{1} (v_{1f})^{2}+m_{2} (v_{2f})^{2}) …………………………………………….(25)
From Eqs. (24) and (25) it follows that,
m_{1}v_{1i} (v_{2f }– v_{1i} ) = m_{1} v_{1f }(v_{2f }– v_{1f })
Or, v_{2f }(v_{1i} – v_{1f })= (v_{1i} )^{2} – (v_{1f })^{2}
= (v_{1i} – v_{1f })((v_{1i} + v_{1f })
Hence, v_{2f }= (v_{1i} – v_{1f }) ………………………………………………………………………..(26)
Substituting this in Eq. (24), we obtain
v_{1f }= [(m_{1}m_{2})/(m_{1}+m_{2})]v_{1i} …………………………………………………………………(27)
v_{2f }= 2m_{1}v_{1i}/(m_{1}+m_{2}) …………………………………………………………………(28)
Thus, the ‘unknowns’ {v_{1f }, v_{2f }} are obtained in terms of the unknowns {m_{1}, m_{2}, v_{1i}}. Special cases of our analysis are interesting.
Case I : If the two masses are equal
v_{1f }= 0
v_{2f }= v_{1f }
The first mass comes to rest and pushes off the second mass with its initial speed on collision.
Case II : If one mass dominates e.g. m_{2 }» m_{1}
v_{1f }≅ v_{1i}
v_{2f }≅ 0
The heavier mass is undisturbed while the lighter mass reverses its velocity.
Example 12
Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically 10^{7 }m s ^{1}) must be slowed to 10^{3 }m s ^{1 }so that it can have a high probability of interacting with isotope and causing it to fission. Show that neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D_{2}O) or graphite, is called a moderator.
Ans.: The initial kinetic energy of the neutron is
K_{1i }= 1/2 m_{1 }(v_{1i})^{2}
while its kinetic energy from Eq. (27)
K_{1f }= 1/2 m_{1 }(v_{1f})^{2}=1/2m_{1 }[(m_{1} – m_{2})/(m_{1} + m_{2})]^{2}(v_{1i})2
The fractional kinetic energy lost is
f_{1 }= K_{1f }/ K_{1i }= [(m_{1} – m_{2})/(m_{1} + m_{2})]^{2}
while the fractional kinetic energy gained by the moderating nuclei K_{2f }/ K_{1i }is
f_{2 }= 1 – f_{1 }(elastic collision)
= (4m_{1}m_{2})/(m_{1 }+ m_{2 })^{2}
One can also verify this result by substituting from Eq. (28).
For deuterium m_{2 }= 2m_{1 }and we obtain f_{1 }= 1/9 while f_{2 }= 8/9. Almost 90% of the neutron’s energy is transferred to deuterium. For the carbon f_{1 }= 71.6% and f_{2 }= 28.4%. In practice, however this number is smaller since headon collisions are rare.
If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one dimensional collision, or headon collision. In the case of small spherical bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 which is at rest. In general, the collision is two dimensional, where the initial velocities and the final velocities lie in a plane.
Collisions in Two Dimensions
Fig. 10 depicts the collision of a moving mass m_{1 }with the stationary mass m_{2}. Linear momentum is conserved in such a collision. Since momentum is a vector this implies three equations for the three directions {x, y, z}. Consider the plane determined by the final velocity directions of m_{1 }and m_{2 }and choose it to be the xy plane. The conservation of the zcomponent of the linear momentum implies that the entire collision is in the xy plane. The x and y component equations are
m_{1 }v_{1i} = m_{1 }v_{1f }cosθ_{1} + m_{2 }v_{2f }cosθ_{2} …………………………………(29)
0 = m_{1 }v_{1f }sin θ_{1} + m_{2 }v_{2f }sinθ_{2} ……………………………………….(30)
One knows {m_{1}, m_{2}, v_{1i}} in most situations. There are thus four unknowns {v_{1f }, v_{2f }, θ_{1} and θ_{2}}, and only two equations. If θ_{1} = θ_{2 }= 0, we regain Eq. (24) for one dimensional collision.
If, further the collision is elastic,
1/2(m_{1 }v_{1i})^{2 }= 1/2(m_{1 }v_{1f})^{2 }+ 1/2(m_{2 }v_{2f})^{2 }……………………………….(31)
We obtain an additional equation. That still leaves us one equation short. At least one of the four unknowns, say θ_{1}, must be made known for the problem to be solvable. For example, θ_{1} can be determined by moving a detector in an angular fashion from the from the x to the y axis. Given {m_{1}, m_{2}, v_{1i}, θ_{1} } we can determine {v_{1f}, v_{2f}, θ_{2}} from equations (29) – (31).
Example 13
Consider the collision depicted in Fig. 10 to be between two billiard balls with equal masses m_{1} = m_{2}. The first ball is called the cue while the second ball is called target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ_{2 }= 37º. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain θ_{1}.
Answer
From momentum conservation, since the masses are equal
v_{1i}= v_{1f }– v_{2f}
or (v_{1i})^{2}= (v_{1f }+ v_{2f}). (v_{1f }+ v_{2f})
=(v_{1f})^{2}+ (v_{2f})^{2}+ 2v_{1f }.v_{2f}
= (v_{1f})^{2}+ (v_{2f})^{2}+ 2v_{1f }v_{2f }cos (θ_{1} + 37º) ………………………………………..(32)
Since the collision is elastic and m_{1 }= m_{2 } it follows from conservation of kinetic energy that,
(v_{1i})^{2}= (v_{1f})^{2} + (v_{2f})^{2}………………………………………………………………………..(33)
Comparing Equations (32) and (33), we get
cos (θ_{1} + 37º) = 0
or θ_{1} + 37º = 90º
Thus, θ_{1} = 53º
This proves the following result: when two equal masses undergo a glancing elastic collision with one of them at rest, after collision, they will move at right angles to each other.
The matter simplifies greatly if we consider spherical masses with smooth surfaces, and assume that collision takes place only when the bodies touch each other. This is what happens in the games of marbles, carrom and billiards.
In our everyday world, collisions take place only when two bodies touch each other. But consider a coming from far distances to the sun, or alpha particle coming towards a nucleus and going away in some direction. Here we have to deal with forces involving action at a distance. Such an event is called scattering. The velocities and directions in which the two particles go away depend on their velocities as well as the type of interaction between them, their masses, shapes and sizes.