Question

29. A block of mass 'M' is placed on the
top of a wedge of mass 4M'. All the
surfaces are frictionless. The system is
released from rest. The distance moved
by the wedge at the instant, the block
reaches the bottom will be
1) 0.2 m
2) 0.4 m
3) 0.8 m
4) Zero
​

Answer 1

Answer:

Option (3) 0.8 m

Explanation:

The horizontal distance travelled by the block of mass M = 4 m

Since there is no force in the horizontal direction

Therefore, the momentum in the horizontal direction will be conserved

If the velocities of block and the wedge in the horizontal direction are

$v_B$ and $v_W$ then

$Mv_B+4Mv_A=0$

if the distance travelled in the horizontal direction by the block and the wedge are $x_B$ and $x_W$ then

$M\frac{dx_B}{dt}+4M\frac{dx_W}{dt}=0$

$\implies Mdx_B+4Mdx_W=0$

$\implies M\Delta x_B+4M\Delta x_W=0$

If the wedge travels distance x then

$\Delta x_W=x$

The block has a displacement of 4 m with respect to the wedge but the wedge also has a displacement of x m in the same time

Therefore, displacement of block with respect to the earth = $4+x$

Thus,

$M\times (4+x)+4M\times x=0$

$\implies 4+x+4x=0$

$\implies 5x=-4$

$\implies x=-0.8$ m

Thus, the wedge travels a distance of 0.8 m

Hope this helps.

Answer 2

The previous answer is absolutely CORRECT. I verified it.