Question

A thin ring of mass $2\text{kg}$ and radius $1\text{m}$ is rolling without slipping on a horizontal plane with velocity $1\text{m/s}$. A small ball of mass $1\text{kg}$ and moving with velocity $2\text{m/s}$ in the opposite direction, hits the ring at a height of $1.8\text{m}$ and goes vertically up with velocity of $1\text{m/s}$. Immediately after the collision, which of the following is true?

• A. Ring undergoes pure rotation about its axis through the centre of mass.
• B. Ring comes to a complete stop.
• C. Friction on the ring by the ground is to the left.
• D. There is no friction between the ring and ground.

Answer 1

Answer: (A), (C)

The correct options are
A Ring undergoes pure rotation about its axis through the centre of mass.
C Friction on the ring by the ground is to the left.

Let us suppose velocity of centre of mass of the ring just after collision is $v$ and angular speed about an axis through its centre of mass is $\omega$.
According to question -


As ring is rolling without slipping before collision, we can infer that there is no friction between the ring and ground, hence no net force acting on the system.
On applying Law of conservation of linear momentum along x - axis - $\overrightarrow{p_i}=\overrightarrow{p_f}$
$\Rightarrow 1\times 2-2\times 1=1\times 0-2\times v$
$\Rightarrow v=0$
Let us take instantaneous axis of rotation at the point of contact P. If suppose friction acts, then its torque will be zero about that point.
On applying Law of conservation of angular momentum.
$\overrightarrow{L_i}=\overrightarrow{L_f}$
Taking anti-clockwise as positive,
$L_i=L_{ball}+L_{ring}=mv{r}_{\perp }+I\omega$
$=-1\times 2\times 1.8+(2\times 2\times 1^2)\times 1$
& $L_f=L_{ball}+L_{ring}$
$=-1\times 1\times 0.6+(2\times 2\times 1^2)\times \omega$
[perpendicular distance of ball from IAOR $=0.6\text{m}$]
$L_i=L_f$
$\Rightarrow -1\times 2\times 1.8+(2\times 2\times 1^2)\times 1=-1\times 1\times 0.6+(2\times 2\times 1^2)\times \omega$
Using$\left[\begin{array}{c}{L}_{ball}=mv{r}_{\perp },L_{ring}=I\omega \\ I=I_{IAOR}=m{r}^2+m{r}^2=2m{r}^2\\ \text{and}v=\omega r\text{for pure rolling}\end{array}\right]$
$\Rightarrow \omega =\frac{1}{4}\text{rad/s}$

As velocity of c.o.m is zero and angular speed about axis passing through c.o.m is $\frac{1}{4}\text{rad/s}$, therefore the ring just after collision is in pure rotation in anti- clockwise direction.
Point of contact i.e bottommost point of the ring has velocity in right direction. Therefore to oppose it, friction will act on the ring in left direction.
Hence, options A and C are correct.