Question

A ball A collides with another identical ball B with velocity $10m/s$ at at an angle $30°$ from the line joining their centers $C_1$ and $C_2$. Select the correct alternative(s):
A) Velocity of ball B after collision is 5m/s
B) Velocity of ball B after collision is $\dfrac{{10}}{{\sqrt 3 }}$ m/s
C) Both the balls move at right angles after collision.
D) Kinetic energy will not be conserved here, because collision is not head on.

Answer 1

Collision is said to be elastic when two bodies collide and do not stick to each other. Law of conservation of momentum will be taken into account for solving the problem as no external force is acting in the total collision process. Using the above two concepts we will bring out the velocity of ball B.

Complete step by step solution:
Let's discuss a few points about elastic collision and conservation of momentum in order to solve the problem.
A collision is said to be an elastic collision if both the kinetic energy and momentum are conserved in the collision. During collision deformity in the bodies takes place and if the collision is perfectly elastic then bodies take their original shape again.
Conservation of momentum: Momentum is neither created nor destroyed but only changed from one form to another through the action of forces. Total momentum of the system remains conserved and constant.
Now we will calculate the velocity of the ball B:
${m_1}, {u_1}$ (mass and initial velocity of ball A)
${m_2}, {u_2}$ (mass and initial velocity of ball B)
${m_1}, {v_1}$ (mass and initial velocity of ball A)
${m_2}, {v_2}$(mass and initial velocity of ball B)
As per conservation of momentum:
${m_1}{u_1}$ = $10m/s$ (We will assume mass of both the balls as 1)
Initial velocity of ball B was zero because it was at rest.
For final momentum, first we will take the x component of the angles which the balls make.
$\Rightarrow {m_1}{v_1}\cos \theta + {m_2}{v_2}\cos \theta$=${m_1}{u_1}$+${m_2}{u_2}$
On substituting the value of angle
$\Rightarrow \dfrac{{\sqrt 3 }}{2}{v_1} + \dfrac{{\sqrt 3 }}{2}{v_2} = 10$
$\Rightarrow {v_1} + {v_2} = \dfrac{{20}}{{\sqrt 3 }}$ .............(1)
If we take the y component of the angles which balls make
$\Rightarrow {m_1}{v_1}\sin \theta - {m_2}{v_2}\sin \theta = 0$
$\Rightarrow \dfrac{{{v_1}}}{2} - \dfrac{{{v_2}}}{2} = 0$
$\Rightarrow {v_1} = {v_2}$ ..............(2)
We will make use of the equation 2 for 1 then we have
$\Rightarrow 2{v_2} = \dfrac{{20}}{{\sqrt 3 }}$(As $v_1$ and $v_2$ are equal)
$\Rightarrow {v_2} = \dfrac{{10}}{{\sqrt 3 }}$ m/s
Thus velocity of ball B is $\dfrac{{10}}{{\sqrt 3 }}$ m/s.
Balls move at right angles after collision as the collision is not head on (colliding bodies do not move in the same straight line).

Thus options B, C and D are correct.

Note: Elastic collision has applications in nuclear reactor when neutrons hits each other for producing more reactions, balls hit each other in the game of billiards, when spacecraft fly in the universe and do not hit the planet, ball thrown on the ground and it returns back.