Q: A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips in the semicircular frictionless track. How far the block moved when the cylinder reaches the bottom of the track?
Sol: Since no net horizontal force acts on the system, it conserves its horizontal momentum.
$\displaystyle \vec{P} = M \vec{v_b} + m \vec{v_c} = 0 $ .
Because initially the system was stationary.
$\displaystyle M \vec{v_b} + m \vec{v_c} = 0 $ , Where $\vec{v_b}$ & $\vec{v_c}$ are the horizontal component of velocities of the block and the cylinder respectively
vcb = velocity of the cylinder relative to the block = v (say)
$\displaystyle v_b = \frac{m v_{cb} sin\theta}{M + m}$ ( Numerically )
$\displaystyle \int_{0}^{t} v_b dt = \frac{m}{M+m}\int_{0}^{t} (v_{cb})_x dt $
where t = time taken by the cylinder to arrive at the bottom of the block& (vcb)x = horizontal component of velocity of the cylinder relative to the block.
$\displaystyle x_b = \frac{m x_{cb}}{M+m} $ ; putting xbc = (R-r) we obtain
$\displaystyle x_b = \frac{m}{M+m}(R-r) $ ; Where $\displaystyle \int_{0}^{t} (v_{cb})_x dt = (R-r)$ and $\displaystyle \int_{0}^{t} v_b dt = x_b $
