Question

A smooth sphere A of mass 'm' moving with
a constant speed 'V' on the smooth
horizontal surface collides elastically with
an identical sphere B at rest. After elastic
collision speed of sphere A is V/2 then the
speed of sphere of B is
​

Answer 1

Given:

A smooth sphere A of mass 'm' moving with a constant speed 'v' on the smooth horizontal surface collides elastically with an identical sphere B at rest. After elastic collision speed of sphere A is v/2.

To find:

Speed of sphere B ?

Calculation:

Since the question clearly mentions elastic collision between the two spheres , we can easily apply CONSERVATION OF LINEAR MOMENTUM principle :

Let final velocity of block B be v_(2):

$\rm \therefore \: P_{initial} = P_{final}$

$\rm \implies \: mv + m(0) = m( \dfrac{v}{2} ) + m (v_{2})$

$\rm \implies \: mv + 0= m( \dfrac{v}{2} ) + m (v_{2})$

$\rm \implies \: mv = \dfrac{mv}{2} + m (v_{2})$

$\rm \implies \:m (v_{2}) = mv - \dfrac{mv}{2}$

$\rm \implies \:m (v_{2}) = \dfrac{mv}{2}$

$\rm \implies \:v_{2}= \dfrac{v}{2}$

So, final velocity of sphere B is v/2 in the same direction as that of sphere A.