Question

Two balls A and B having masses 1Kg and 2Kg, moving with speeds 21m/s and 4m/s respectively in opposite directions, collide head on. After collision A moves with a speed of 1m/s in the same direction, the correct statement is:

(This question has multiple correct options)

$(a)$ The velocity of B after collision is 6m/s opposite to the direction of the motion before collision

$(b)$ The coefficient of restitution is 0.2

$(c)$ The loss of kinetic energy due to collision is 200J

$(d)$ The impulse of the force between the two balls is 40Ns

Answer 1

Hint: In this question take the velocities along the + x axis as positive while velocities in negative x axis as negative. So velocity of ball A be 21m/s and that of B be -4m/s before collision. After collision the velocity changes thus let the velocity of the ball B after collision  be ${v_2}$ m/s. Application of conservation of momentum will help getting the answer. 

Complete step by step answer:

Given data:

Before colliding, let us assume ball A moves in positive direction of x-axis therefore ball B moves in negative direction of x-axis (as they moves in opposite direction)

Ball A: ${m_1} = 1$ Kg and ${u_1} = 21$ m/s

Ball B: ${m_2} = 2$ Kg and ${u_2} =  - 4$ m/s (‘-’ sign indicates negative direction of x-axis)

After colliding

Ball A: ${m_1} = 1$ Kg and ${v_1} = 1$ m/s (continue moving in same direction)

Ball B: ${m_2} = 2$ Kg and ${v_2} = $? m/s

It is given that they are moving in the opposite direction and collide head on. After colliding, ball A moves in the same direction as shown in the figure.

Now according to conservation of momentum, momentum before colliding is equal to the momentum after colliding.

And we all know momentum is the product of mass and velocity.

So, the momentum will be conserved. 

Therefore, sum of momentum of ball A and ball B before collision = sum of momentum of ball A and ball B after collision.

$ \Rightarrow {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Now substitute the values we have,

$ \Rightarrow \left( 1 \right)\left( {21} \right) + \left( 2 \right)\left( { - 4} \right) = \left( 1 \right)\left( 1 \right) + \left( 2 \right){v_2}$

Now simplify this we have,

$ \Rightarrow 21 - 8 = 1 + 2{v_2}$

$ \Rightarrow 2{v_2} = 13 - 1 = 12$

$ \Rightarrow {v_2} = \dfrac{{12}}{2} = 6$ m/s.

So as the velocity of ball B after colliding comes positive so ball B moves in the positive direction of x-axis. Therefore, The velocity of B after collision is opposite to the direction of the motion before collision .Hence option (A) is a correct answer.

Now the coefficient of restitution (e) is the magnitude of the ratio of difference of velocity after collision to difference of velocity  before collision therefore,

$ \Rightarrow e = \left| {\dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}}} \right|$

Now substitute the values we have,

$ \Rightarrow e = \left| {\dfrac{{1 - 6}}{{21 - \left( { - 4} \right)}}} \right| = \left| {\dfrac{{ - 5}}{{25}}} \right| = \dfrac{5}{{25}} = \dfrac{1}{5} = 0.2$

Hence option (B) is also a correct answer.

As we know that the kinetic energy gained by the ball is equal to $\dfrac{1}{2}m{v^2}$

So the kinetic energy lost by ball A is the difference of the kinetic energy of ball A before the collision and after the collision, same as for ball B.

So the total kinetic energy lost = kinetic energy lost by ball A + kinetic energy lost by ball B

Therefore, kinetic energy lost = $\dfrac{1}{2}{m_1}\left( {u_1^2 - v_1^2} \right) + \dfrac{1}{2}{m_2}\left( {u_2^2 - v_2^2} \right)$

Now substitute the values we have,

$ \Rightarrow \dfrac{1}{2}\left( 1 \right)\left( {{{\left( {21} \right)}^2} - {{\left( 1 \right)}^2}} \right) + \dfrac{1}{2}\left( 2 \right)\left( {{{\left( { - 4} \right)}^2} - {{\left( 6 \right)}^2}} \right)$

Now simplify this we have,

$ \Rightarrow \dfrac{1}{2}\left( {441 - 1} \right) + \left( {16 - 36} \right) = \dfrac{{440}}{2} - 20 = 220 - 20 = 200$J

Hence option (C) is also the correct answer.

Impulse of force on ball A is the product of mass and change in velocity, same as for ball B.

So impulse of force (I.F) between the two balls is the difference of the impulse of force of the respective balls. The magnitude of  impulse of  force acting on ball A is equal to the magnitude of the impulse of force acting on ball B but in the opposite direction. So, to find the impulse of force acting between the two balls, we only have to find the impulse of force ball A or ball B. Here, we are calculating the impulse of force on ball A.

Therefore, I.F = ${m_1}\left( {{v_1} - {u_1}} \right) $

Now substitute the values we have,

I.F = $\left( 1 \right)\left( {21 - 1} \right)$

I.F = $20$ kgm/s (or) N-s

Hence option (D) is a wrong answer.

Hence all the options A,B, and C are correct.

Note: The coefficient of restitution is always in the range of 0 to 1 that is $0 \leqslant e \leqslant 1$. In case of a perfectly inelastic collision the coefficient of restitution is 0, this means that the two balls will stick together after the collision has taken place. The total kinetic energy loss of the entire system is equal to the losses due to the individual components of that system.