Question

A block of mass M$=2$ kg with a semicircular track of radius R$=1.1$ m rests on a horizontal frictionless surface. A uniform cylinder of radius $r=10$ cm and mass m$=1.0$ kg is released from rest from the top point A. The cylinder slips on the semicircular frictionless track. The speed of the block when the cylinder reaches the bottom of the track at B is : $(g=10m/s^2)$

• A. $\sqrt{\frac{10}{3}}$m/s
• B. $\sqrt{\frac{4}{3}}$m/s
• C. $\sqrt{\frac{5}{2}}$m/s
• D. $\sqrt{10}$m/s

Answer 1

The correct option is A $\sqrt{\frac{10}{3}}$m/s

Momentum is conserved for the system

$P_m=P_M$ ( In opposite directions )

$m{V}_1=(m+M)V_2$

$K.E=\frac{1}{2}m{v}^2$ $(M_1=m+M)$

So,

$\sqrt{2KmM}=\sqrt{2KmM}$

$\frac{Km}{K{M}_1}=\frac{M_1}{m}$

$=K{M}_1=\frac{m}{(m+M)}\times Km$ ---------------(I)

Now kinetic energy will be changed in potential energy

Change in potential energy $=mg(r-R)$ ------(II)

Using equation (I) and (ii)

$\frac{1}{2}M{V}^2=\frac{m}{(m+m)}\times \left(mg\right(r-R\left)\right)$

$V^2=\frac{2m^{2}g(r-R)}{M(m+M)}$

$V^2=2\times 10\times 1.1-\frac{0.1}{2\times 3}$

$⟹V=\sqrt{\frac{10}{3}}m/s$