Question

A block of mass $m$ is moving with velocity $u$ as shown in the figure. The wedge has mass $nm$ and height $h$. All surfaces are smooth.

• A. Velocity of centre of mass is zero when block reaches the highest point on the wedge
• B. Horizontal component of velocity of centre of mass is $\frac{u}{1+n}$
• C. If the block has to reach the top of wedge then $u=\sqrt{2gh(1+\frac{1}{n^3})}$
• D. Block reaches top of the wedge when velocity $u=\sqrt{0.24gh}$

Answer 1

The correct option is B Horizontal component of velocity of centre of mass is $\frac{u}{1+n}$

Velocity of center pf mass of wedge and block system is $v_{com}=\frac{nm\times 0+mu}{nm+m}=\frac{mu}{(n+1)m}$

$v_{com}=\frac{u}{n+1}$