Question

A block of mass $m_1$ is pushed towards the movable wedge of mass $m_2$ and height $h$, with a velocity $v_0$. All surfaces are smooth. The minimum value of $v_0$ for which the block will reach the top of the wedge is

• A. $h=\frac{v_0^2}{4(1+\frac{m_1}{m_2})g}$
• B. $h=\frac{v_0^2}{2(1+\frac{m_2}{m_1})g}$
• C. $h=\frac{3v_0^2}{2(1+\frac{m_1}{m_2})g}$
• D. $h=\frac{v_0^2}{2(1+\frac{m_1}{m_2})g}$

Answer 1

The correct option is D $h=\frac{v_0^2}{2(1+\frac{m_1}{m_2})g}$


Using Work energy theorem in COM reference frame,
$K_i+U_i=K_f+U_f$
Here , $K_i=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{v}_{rel}^2=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{v}_0^2$

Assuming initial position as reference for potential energy
$U_i=0$
Let $h$ be maximum height reached by $m_1$ on $m_2$
So, $U_f=m_{1}gh$
For $v_o$ to be minimum, velocity of $m_1$ relative to $m_2$ will be zero (when $m_1$ reaches height $h$ on $m_2$)
$K_f=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{v}_{rel}^2=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{0}^2=0$

$\therefore$
$m_{1}gh=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{v}_0^2$
$h=\frac{m_2{v}_0^2}{2g(m_1+m_2)}$
$h=\frac{v_0^2}{2(1+\frac{m_1}{m_2})g}$