Question

A ball of mass m is allowed to role down the wedge of mass M $=$ 2m as shown in the figure. What is the displacement of wedge (in m) when the ball reaches from A to B? Take $\theta ={45}^o,h=1m,d=4m$

• A. $4m$
• B. $3m$
• C. $2m$
• D. $5m$

Answer 1

The correct option is C $2m$

As there is no force acting on the system in horizontal direction.

So, the horizontal component of COM does not change .

Assuming the point below A is zero.

Let the Wedge move by a distance $x$

A/c to conservation of position of COM in horizontal direction

$M\times (\frac{d}{2}+\frac{h}{tan\theta })+m\times 0$= $M\times (x+\frac{d}{2}+\frac{h}{tan\theta })+m\times (d+x+\frac{2h}{tan\theta })$

$⟹$ $M\times x+m\times (d+x+\frac{2h}{tan\theta })=0$

$⟹$ $x=\frac{-md-\frac{2h}{tan\theta }}{M+m}$

$⟹$ $x=\frac{-m\times 4-\frac{2times1}{tan{45}^o}}{2m+m}$

$⟹$ $x=2metre$