Question

A wooden block is cut into two pieces; one is 3 times heavier than the other. A depression is made in bothpieces; a firecracker is inserted in the depression and the block is reassembled. The reassembled block is placedon a rough surface and the fuse on the firecracker is lit. When the firecracker explodes, the two pieces of theblock slide apart. What is theratioof the distances traveled by each block?

Answer 1

The ratio of their distance is 9:1

Solution:

As per the question:

Suppose,

Mass of the lighter block = m

Mass of the heavier block, M = 3m

Velocity of the lighter block be v and the velocity of the heavier block be v'.

Now,

The initial momentum of the system is zero.

Therefore,

mv + Mv' = 0

mv = - Mv'

mv = - 3mv'

`(v)/(v') = -3` (1)

Now, if the distance covered by the lighter and the heavier block be d and d' respectively, then by the kinematic eqn:

`v^(2) = 2ad` (2)

`v'^(2) = 2ad'` (3)

Dividing eqn (2) by (3), we get:

`((v)/(v')^(2)) = (d)/(d')`

`(d)/(d') = (-3)^(2)`

`(d)/(d') = 9`