Question

A solid sphere of mass $m=80\text{kg}$ and radius $r=0.2\text{m}$ is released from height $h=\frac{5}{4}$ meter. Sphere is initially rotating about horizontal axis passing through its centre of mass. It hits with a stationary cart of mass $M=200\text{kg}$ exactly at the centre of cart. The cart can move smoothly on the horizontal surface.
The collision between sphere and cart occurs in such a way that sphere reaches at same vertical displacement after collision and falls back onto it again. It is found that sphere starts pure rolling at the end of first collision. The coefficient of friction between sphere and cart is $\mathrm{0.1.}$
Match the statement given in List-I to the values(in SI units) given in List-II.

$\begin{array}{cccc}& \text{Column-I}& & \text{Column-II}\\ \text{P.}& \text{The minimum length (in meter) of cart to occur second collision with the sphere.}& 1.& 172\\ \text{Q.}& \text{Initial angular velocity}{}^{\prime }{\omega }^{\prime }\left(\text{in}\frac{\text{rad}}{\text{sec}}\right)\text{of sphere on the cart during the process.}& 2.& 2.8\\ \text{R.}& \text{Magnitude of work done (in Joule) by sphere on the cart during the process.}& 3.& 156\\ \text{S.}& \text{Magnitude of work done (in Joule) by cart on the sphere during the process}& 4.& 19.5\\ & & 5.& 16\end{array}$
Which of the following option has the correct combination considering column –I and column – II

• A. $S\to 3$
• B. $Q\to 4$
• C. $R\to 2$
• D. $P\to 1$

Answer 1

The correct option is B $Q\to 4$

Final vertical velocity of sphere after it released
$\Rightarrow v_0=\sqrt{2gh}=5\frac{\text{m}}{\text{s}}....\left(1\right)$
Now, impulse in y direction
$\Rightarrow J_y=\Delta P_y=2\times 80\times 5=800\text{Ns}...\left(2\right)$

Similarly, impulse in x direction
$\Rightarrow J_x=\Delta P_x$
$\Rightarrow \mu J_y=m{v}_x$
$\Rightarrow v_x=1{\text{ms}}^{-1}...\left(iii\right)$
Impulse acting on cart is
$\Rightarrow M{v}_c=J_x$
$\Rightarrow v_c=0.4{\text{ms}}^{-1}......\left(iv\right)$
time interval between first and second collision is
$\Rightarrow t_0=\frac{2\times 5}{10}=1\text{sec}...\left(v\right)$
$\Rightarrow L_{min}=u{t}_0=2(1+0.4)1=2.8\text{m}...\left(vi\right)$
$\Rightarrow$ At the time of second collision for pure rolling
$\Rightarrow R{\omega }^{\prime }-v_x=v_c$
$\Rightarrow 0.2{\omega }^{\prime }=1.4$
$\Rightarrow {\omega }^{\prime }=7\frac{\text{rad}}{\text{s}}....\left(vii\right)$

Angular impulse on sphere is
$\Rightarrow -J_{x}R=I({\omega }^{\prime }-{\omega }_0)$
$\Rightarrow -16=\frac{2}{5}\left(80\right)\left(0.2{)}^2\right(7-{\omega }_0)$
$\Rightarrow {\omega }_0=19.5\frac{\text{rad}}{\text{sec}}...\left(viii\right)$
Work done by sphere on the cart
$\Rightarrow W_{m\to M}=\frac{1}{2}M{v}_c^2=16\text{J}$
Work done by cart on the sphere is
$\Rightarrow W_{M\to m}=\frac{1}{2}I({\omega }^{\prime }{)}^2+\frac{1}{2}m{v}_x^2-\frac{1}{2}I{\omega }_0^2=-172\text{J}$