Question

A body moving along a straight line collides another body of same mass moving in the same direction with half of the velocity of the first body. If the coefficient of restitution between the two bodies is 0.5, then the ratio of the velocities of the two bodies after collision is (treat the collision as one dimensional)

• A. 2:5
• B. 2:3
• C. 5:7
• D. 3:7

Hint:

Coefficient of restitution: $e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$. Conservation of momentum.

Answer 1

The Correct Option is C

Let the masses of the bodies be $m$. Let the initial velocities be $u_1$ and $u_2 = \frac{1}{2}u_1$. Let the final velocities be $v_1$ and $v_2$. Coefficient of restitution $e = \frac{v_2 - v_1}{u_1 - u_2} = 0.5$. Since the masses are equal, we can use the following simplified equations for one-dimensional elastic collisions: $v_1 = \frac{u_1(m-em)+u_2 m(1+e)}{2m} = \frac{u_1+u_2+e(u_2-u_1)}{2}$ and $v_2 = \frac{u_1+u_2+e(u_1-u_2)}{2}$ If we use $u_2=\frac{u_1}{2}$ and $e=0.5$, then $v_2-v_1 = 0.5(u_1-\frac{u_1}{2})=0.5\frac{u_1}{2} = \frac{u_1}{4}$. Using Conservation of momentum: $mu_1+\frac{mu_1}{2} = mv_1+mv_2$ so $\frac{3u_1}{2}=v_1+v_2$. $v_2=v_1+\frac{u_1}{4}$. So $\frac{3u_1}{2}=2v_1+\frac{u_1}{4}$ so $2v_1=\frac{5u_1}{4}$ so $v_1=\frac{5u_1}{8}$ and $v_2=\frac{7u_1}{8}$. $v_2 - v_1 = \frac{1}{2}(u_1 - \frac{1}{2}u_1) = \frac{1}{4}u_1$ Also, from conservation of momentum, $m(u_1 + \frac{1}{2}u_1) = m(v_1 + v_2)$, so $v_1 + v_2 = \frac{3}{2}u_1$. Solving for $v_1$ and $v_2$, we get $v_1 = \frac{5}{8}u_1$ and $v_2 = \frac{7}{8}u_1$. The ratio of velocities is $\frac{v_1}{v_2} = \frac{5/8}{7/8} = \frac{5}{7}$.