Question

A block of mass $2\text{kg}$ is projected with a velocity $6\text{m/s}$ so that it climbs onto z smooth wedge of mass $1\text{kg}$. If the block does not leave the wedge, find the maximum height attained by the block. Take $g=10{\text{m/s}}^2$.

• A. $0.3\text{m}$
• B. $0.6\text{m}$
• C. $1.2\text{m}$
• D. $0.4\text{m}$

Answer 1

The correct option is B $0.6\text{m}$

The block attains a maximum height when the final relative velocity between the block and wedge becomes zero. Final velocity of wedge is equal to final velocity of the block.

Kinetic energy of the system with respect to $COM$ is given by
$K_{com}=\frac{1}{2}\frac{m_1{m}_2}{(m_1+m_2)}(v_1-v_2{)}^2$

Final kinetic energy of system with respect to $COM$ is
$(K_{\text{com}}{)}_{\text{final}}=\frac{1}{2}\frac{m_1{m}_2}{(m_1+m_2)}(v_{1f}-v_{2f}{)}^2$
$(K_{\text{com}}{)}_{\text{final}}=0$ [$\because v_{1f}=v_{2f}]$

and Initial kinetic energy of system with respect to $COM$ is
$(K_{\text{com}}{)}_{\text{initial}}=\frac{1}{2}\frac{2\times 1}{3}(6-0{)}^2=12\text{J}$

Let $h$ be the height reached by block.
Using work energy theorem in $COM$ frame:
$\Delta W_{net}=\Delta KE$
$-m_{1}gh=(K_{\text{com}}{)}_{\text{final}}-(K_{\text{com}}{)}_{\text{initial}}$
$-2\times 10\times h=0-12$
$h=\frac{12}{2\times 10}$
$\therefore h=0.6\text{m}$