Question

A ball moving with velocity 2m/s collides head on with another stationary ball of double the mass. If coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be:
A. 0,1
B. 1,1
C. 1,0,5
D. 0,2

Answer 1

Hint: To solve this question, we need to use the basic theory related to the chapter collision. As we know when two particles collide with each other then velocity of these two after collision may be changed based on the type of the collision (Elastic or inelastic collision). Since, in this question the coefficient of restitution is 0.5. This is a real-world inelastic collision, in which some kinetic energy is dissipated. And we use the formula for the velocities after a one-dimensional collision when the coefficient of restitution is given.

Formula used: Velocity of first ball after collision $v_1={\displaystyle \frac{\left(m_{1-}e{m}_2\right)u_1+(1+e)m_2{u}_2}{m_1+m_2}}$.
  Velocity of second ball after collision $v_2={\displaystyle \frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}}$

 Complete step-by-step solution -
Let mass of first ball be m i.e.$m_1=m$
Mass of second ball$m_2=2m$
velocity of first ball before collision $u_1=2m/s$
velocity of first ball before collision $u_1=2m/s$
Velocity of second ball before collision $u_2=0m/s$
Given: e=0.5
So, velocity of first ball after collision $v_1={\displaystyle \frac{\left(m_{1-}e{m}_2\right)u_1+(1+e)m_2{u}_2}{m_1+m_2}}$.
$\therefore v_1={\displaystyle \frac{\left[m-0.5(2m)\right](2)+(1+0.5)(2m)(0)}{m+2m}}=0m/s$.
Velocity of second ball after collision $v_2={\displaystyle \frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}}$.
$\therefore v_2={\displaystyle \frac{\left[2m-0.5m\right](0)+(1+0.5)(m)(2)}{m+2m}}=1m/s$.
Thus, the velocity of the first ball after collision is 0 m/s and Velocity of the second ball after collision is 1 m/s.
Therefore, Correct answer is option (A).

Note: Consider particles 1 and 2 with masses ${\text{m}}_{\text{1}}$, ${\text{m}}_2$, and velocities ${\text{u}}_{\text{1}}$, ${\text{u}}_{\text{2}}$ before collision, ${\text{v}}_{\text{1}}$, ${\text{v}}_2$ after collision. The conservation of the total momentum before and after the collision is expressed as shown below:
                                                            ${\text{m}}_{\text{1}}{\text{u}}_{\text{1}}\text{ + }{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}\text{ = }{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{ + }{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}$