Question

A small particle of mass m hits a spherical ball of mass M at speed of $V_0$ as shown in the figure. The coefficient of restitution between particle and sphere is 1 and that between sphere and horizontal surface is 0. Then, (given$\frac{M}{m}=3$ and $\theta ={37}^{\circ }$)

• A. speed of sphere after collision is $\frac{24V_0}{91}$.
• B. impulse on particle by ball is $\frac{90m{V}_0}{91}$.
• C. speed of particle after collision is equal to $V_0$.
• D. speed of sphere after collision is $\frac{37V_0}{91}$.

Answer 1

Answer: (A), (B)

The correct options are
A speed of sphere after collision is $\frac{24V_0}{91}$.
B impulse on particle by ball is $\frac{90m{V}_0}{91}$.


$J=m[V_1+V_0\sin (\theta \left)\right]................\left(1\right)J\cos \left(\theta \right)=M{V}_x................\left(2\right)e=1=\frac{V_1+V_x\cos \left(\theta \right)}{V_0\sin \left(\theta \right)}\Rightarrow V_0\sin \left(\theta \right)=V_1+V_x\cos \left(\theta \right)................\left(3\right)Solving\left(1\right),\left(2\right)and\left(3\right)V_x=\frac{24V_0}{91}and{V}_1=\frac{177V_0}{455}J=m[\frac{177V_0}{455}+\frac{3V_0}{5}]=\frac{450m{V}_0}{455}=\frac{90m{V}_0}{91}$