Question

A spherical shell of mass M and radius R is released from height h on smooth inclined plane AB having angle of inclination ${37}^{\circ }$ with horizontal. Ball comes down and strikes horizontal rough plane such that immediately after collision vertical velocity of center of mass becomes zero and it moves parallel to ground. Coefficient of friction between ground and spherical body is $\mu =0.5$.
Velocity and angular velocity of body immediately after collision are 'V' and $w$. Component of impulse due to ground during collision are $I_x$ (horizontal) and $I_y$ (vertical) and after collision body is doing pure rolling. $(g=10/,m/s^2)$

Match variable in List - I with List - 2.

$\begin{array}{cccc}& \text{List-1}& & \text{List-2}\\ \text{(I)}& \frac{V}{\sqrt{2gh}}& \text{(P)}& \frac{3}{4}\\ \text{(II)}& \frac{wR}{\sqrt{2gh}}& \text{(Q)}& \frac{3}{5}\\ \text{(III)}& \frac{I_y}{M\sqrt{2gh}}& \text{(R)}& \frac{4}{7}\\ \text{(IV)}& \frac{I_x}{M\sqrt{2gh}}& \text{(S)}& \frac{8}{25}\\ & & \text{(T)}& \frac{8}{35}\\ & & \text{(U)}& \frac{12}{25}\end{array}$

• A. I $\to$ U, II $\to$ U, III $\to$ Q, IV $\to$ S
• B. I $\to$ R, II $\to$ R, III $\to$ S, IV $\to$ P
• C. I $\to$ P, II $\to$ P, III $\to$ Q, IV $\to$ S
• D. I $\to$ U, II $\to$ U, III $\to$ S, IV $\to$ T

Answer 1

The correct option is A I $\to$ U, II $\to$ U, III $\to$ Q, IV $\to$ S


$V_0=\sqrt{2gh}$

$V_x=V_{0}cos\theta$

$V_y=V_{0}sin\theta$

$I_y=\Delta P_y\Rightarrow I_y=M{V}_y=M{V}_{0}sin\theta$

$I_y=M\sqrt{2gh}\times \frac{3}{5}\Rightarrow \frac{I_y}{M\sqrt{2gh}}=\frac{3}{5}$

$I_x=\Delta P_x=M{V}_x-MV$

$I_x=M[V_{0}cos\theta -V]\Rightarrow \frac{I_x}{M}=V_0\times \frac{4}{5}-V\dots \dots \left(1\right)$

Also $I_{x}R=Iw$ also $V=Rw$

$I_x=\frac{I}{R^2}V\dots \dots \left(2\right)$

By (1) and (2)

$\frac{I_x}{M}=\frac{4}{5}{V}_0-I_x\frac{R^2}{I}\Rightarrow \frac{I_x}{M}+I_x\frac{R^2}{I}=\frac{4}{5}{V}_0$

For spherical shell
$I=\frac{2}{3}M{R}^2$

So, $\frac{I_x}{M}+\frac{3}{2}\frac{I_x}{M}=\frac{4}{5}{V}_0\Rightarrow \frac{5}{2}\frac{I_x}{M}=\frac{4}{5}{V}_0$

$\frac{I_x}{M\sqrt{2gh}}=\frac{8}{25}$ also $\frac{I_x}{M}=\frac{8}{25}{V}_0$

So by (1)

$\frac{8}{25}{V}_0=\frac{4}{5}{V}_0-V\Rightarrow \frac{V}{\sqrt{2gh}}=(\frac{4}{5}-\frac{8}{25})$

$\frac{V}{\sqrt{2gh}}=\frac{12}{25}$

$\frac{Rw}{\sqrt{2gh}}=\frac{12}{25}$