Question

A block of mass $2\text{kg}$ with a semicircular track of radius $5\text{m}$, rests on a horizontal frictionless surface. A uniform cylinder of radius $1\text{m}$ and mass $1\text{kg}$ is released from rest at the point A. If the cylinder slips on the semicircular frictionless track, then which of the following is/are correct?

• A. Block moves by $\frac{2}{3}\text{m}$ when the cylinder reaches point B.
• B. Block moves by $\frac{4}{3}\text{m}$ when the cylinder reaches point B.
• C. Speed of the block when the cylinder reaches point B is $\frac{3\sqrt{20}}{5}\text{m/s}$.
• D. Speed of the block when the cylinder reaches point B is $\frac{2\sqrt{30}}{3}\text{m/s}$.

Answer 1

Answer: (B), (D)

The correct options are
B Block moves by $\frac{4}{3}\text{m}$ when the cylinder reaches point B.
D Speed of the block when the cylinder reaches point B is $\frac{2\sqrt{30}}{3}\text{m/s}$.

Given,
Mass of block $\left(m_1\right)=2\text{kg}$
Mass of the cylinder ($m_2)=1\text{kg}$
Radius of semi circular track $\left(r_1\right)=5\text{m}$
Radius of cylinder ($r_2)=1\text{m}$
Let us consider the ball and block as one system.
From the data given in the question, it is clear that there is no net force acting along the horizontal direction.
$\therefore$ Centre of mass of the system will not change in horizontal direction.
From the law of conservation of momentum we can say that, if cylinder move towards right then to maintain same position of centre of mass in horizontal direction, block will move to its left.


For centre of mass to be at rest along the horizontal, we apply
$m_2{d}_2=m_1{d}_1$
where, $d_1$ and $d_2$ are the distances moved by the block and cylinder along horizontal direction.
From the figure, we can write that
$2\times d=1\times (4-d)$
$\Rightarrow d=\frac{4}{3}\text{m}$
$\therefore$ We can deduce that the block has moved $\frac{4}{3}\text{m}$ towards left to keep the centre of mass at rest.
Let $v_1$ and $v_2$ be the final speeds of the block and cylinder.
On applying law of conservation of linear momentum,
$\overrightarrow{p_i}=\overrightarrow{p_f}$
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$\Rightarrow 0+0=2\times v_1-1\times v_2$
$\Rightarrow v_2=2v_1.............\left(1\right)$
On applying Law of conservation of mechanical energy,
$M.E_i=M.E_f$
$\frac{1}{2}{m}_1(u_1{)}^2+\frac{1}{2}{m}_2(u_2{)}^2+m_2\times g\times h_1=\frac{1}{2}{m}_1(v_1{)}^2+\frac{1}{2}{m}_2(v_2{)}^2+m_2\times g\times h_2$
where $h_1$ and $h_2$ are the positions of centre of mass of cylinder from the reference level.
$\Rightarrow 0+0+1\times 10\times 4=\frac{1}{2}\times 2\times v_1^2+\frac{1}{2}\times 1\times v_2^2+0$
Using $\left(1\right)$ in the above equation we get, $4v_1^2+2v_1^2=80\text{m/s}$
$\Rightarrow v_1=\sqrt{\frac{80}{6}}$
$\Rightarrow$ Speed of block $\left(v_1\right)=\frac{2\sqrt{30}}{3}\text{m/s}$
Hence, options (b) and (d) are correct answers.