Question

A ball collides elastically with the same mass of the other ball. Initially one of the balls was at rest and the collision was oblique. After that, the collision of the two balls moves at the same speeds. What will be the angle between the two velocities of the balls after the collision?
$(A){30^ \circ }$
$(B){45^ \circ }$
$(C){60^ \circ }$
$(D){90^ \circ }$

Answer 1

Conservation of momentum is applicable in describing collisions between objects. The law of conservation of momentum is defined as the two or more bodies in an isolated system applying upon each other; their total momentum remains constant unless an external force is applied which can neither be created nor destroyed. We can find the angle between the velocities of the balls after the collision by applying the law of conservation of momentum.

Complete step by step solution:
A ball collides with the second ball of the same mass $m$. The velocities of the two balls are $u\& v$. The angles between the velocities of the balls are ${\theta _1}\& {\theta _2}$.
Applying the law of conservation of momentum along the direction of initial velocity we get,
$\Rightarrow mu = mv(\cos {\theta _1} + \cos {\theta _2})$
Applying the law of conservation of momentum perpendicular to the direction of initial velocity,
$\Rightarrow \theta = mv\sin {\theta _1} - mv\sin {\theta _2}$
$\Rightarrow {\theta _1} = {\theta _2}$
$\Rightarrow mu = 2mv\cos \theta \left[ {{\theta _1} = {\theta _2} = \theta } \right]$
$\Rightarrow \cos \theta = \dfrac{u}{{2v}} \to \left( 1 \right)$
According to the law of conservation of KE,
$\Rightarrow \dfrac{1}{2}m{u^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2}$
$\Rightarrow {u^2} = 2{v^2}$ Or $u = \sqrt 2 v \to \left( 2 \right)$
From the equation $\left( 1 \right)\& \left( 2 \right)$, we have $\cos \theta = \dfrac{{\sqrt 2 v}}{{2v}}$
Or $\cos \theta = \dfrac{1}{{\sqrt 2 }}$
Or $\theta = {45^ \circ }$
${\theta _1} + {\theta _2} = {90^ \circ }$
Hence, the correct answer is the angle between the velocities of the balls after the collision is ${90^ \circ }$.

Note: The mass of an object multiplied is equal to the momentum by its velocity and is equivalent to the force required to bring the object to a stop in a unit length of time.
When the collision of two particles, the sum of the two momenta before the collision is equal to their sum after the collision.