Solveeit Logo

Question

Physics Question on work, energy and power

A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution is

A

122\frac{1}{2\sqrt{2}}

B

13\frac{1}{\sqrt{3}}

C

12\frac{1}{\sqrt{2}}

D

32\frac{\sqrt{3}}{2}

Answer

12\frac{1}{\sqrt{2}}

Explanation

Solution

Let u and v be the initial and final velocities of ball 1 and u and v be the similar quantities for ball 2. Here, u = 0 and v = 0. \therefore\quad initial KE, Ki=12mu12+12mu22=12mu12K_{i}=\frac{1}{2}mu^{2}_{1}+\frac{1}{2}mu^{2}_{2}=\frac{1}{2}mu^{2}_{1} and final KE, Kf=12mv12+12mv22=12mv22K_{f} =\frac{1}{2}mv^{2}_{1}+\frac{1}{2}mv^{2}_{2}=\frac{1}{2}mv^{2}_{2} Loss of KE, ΔK=KiKf=12mu1212mv22\Delta K=K_{i}-K_{f} =\frac{1}{2}mu^{2}_{1}-\frac{1}{2}mv^{2}_{2} According to question, 12(12mu12)=12mu1212mv22\frac{1}{2}\left(\frac{1}{2}mu^{2}_{1}\right)=\frac{1}{2}mu^{2}_{1}-\frac{1}{2}mv^{2}_{2} (\because \, half of its KE is lost by impact) oru12=2v22\quad u^{2}_{1}=2v^{2}_{2} or v2=u12v_{2}=\frac{u_{1}}{\sqrt{2}} \therefore\quad Coefficient of restitution, e=v2v1u1u2=v2u1=12e=\left|\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\right|=\frac{v_{2}}{u_{1}}=\frac{1}{\sqrt{2}}