Question

The inclined surfaces of two move able wedges of the same mass M are smoothly conjugated with the horizontal plane as shown in the figure. A small block of mass 'm' slides down the left wedge from a height 'h'. To what maximum height will the block rise along the right wedge?
Neglect the friction

• A. $\frac{h}{(M+m{)}^2}$
• B. $\frac{hM}{(M+m{)}^2}$
• C. $h{\left(\frac{M}{M+m}\right)}^2$
• D. $h\left(\frac{M}{M+m}\right)$

Answer 1

The correct option is C $h{\left(\frac{M}{M+m}\right)}^2$

Since there is no external force , we ca use energy conservation and momentum conservation.

Applying energy conservation for left wedge:

$\frac{m{u}^2}{2}+\frac{m{v}^2}{2}=mg{h}_{max}$

applying momentum conservation for motion on left wedge:

$Mv=mu$

Applying energy conservation for right wedge

$\frac{m{u}^2}{2}=mg{h}_{max}+\frac{1}{2}(M+m)V^2$

$(M+m)V=mu$

solving for $h_{max}$:

$(M+m{)}^2{h}_{max}=h{M}^2$

so $h_{max}=h\frac{M^2}{(M+m{)}^2}$

Hence option C is correct.