Question

A particle of mass 'm' moving with velocity 'v' collides with a stationary particle of mass 2m. After collision the speed of the stationary particle will be

Answer 1

Answer: After collision the speed of the stationary particle  will be $v' = \dfrac{v}{3}$

Explanation:

Given that,

Mass of the particle $m_{1}=m$

Velocity of the moving particle $u_{1} = v$

Mass of the stationary particle $m_{2}=2m$

Velocity of stationary particle $u_{2}= 0$

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}= (m_{1}+m_{2})v'$

$mv+0=(m+2m)v'$

$v' = \dfrac{v}{3}$

Hence, After collision the speed of the stationary particle  will be $v' = \dfrac{v}{3}$

Answer 2

"According to the data given, Conservation of momentum is to be applied in the given situation and therefore, initial momentum before collision will be equal to final momentum after collision. Thereby,

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 mv_2$

$\Rightarrow m \times v + 2m \times 0 = ( m + 2m) \times v$'

$\Rightarrow m v = 3 m v$'

$\Rightarrow v'm v / 3 m$

$\Rightarrow v' = v / 3.$"