Question

A spherical soild 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. $(g=10m/s^2)$

Match variables in column I with column II if spherical body is solid sphere.

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

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

Answer 1

The correct option is B I $\to$ R, II $\to$ R, III $\to$ Q, IV $\to$ T

I $\to$ R, II $\to$ R, III $\to$ Q, IV $\to$ T

$V_0=\sqrt{2gh}$

$V_x=V_{0}cos\theta$

$V_y=V_{0}sin\theta$

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

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

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

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

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 solid sphere $I=\frac{2}{5}M{R}^2$

So $\frac{I_x}{M}+\frac{5I_x}{2M}=\frac{4}{5}{V}_0\Rightarrow \frac{7}{2}\frac{I_x}{M}=\frac{4}{5}{V}_0$

$\frac{I_x}{M\sqrt{2gh}}=\frac{8}{35}$ also $\frac{I_x}{M}=\frac{8}{35}{V}_0$

So by (1)

$\frac{8}{35}{V}_0=\frac{4}{5}{V}_0-V\Rightarrow V=(\frac{4}{5}-\frac{8}{35})V_0$

$\frac{V}{\sqrt{2gh}}=\frac{4}{7}$ and $\frac{Rw}{\sqrt{2gh}}=\frac{4}{7}$ since $V=Rw$