Answer:
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'
[tex]\frac{v}{v'} = -3[/tex] (1)
Now, if the distance covered by the lighter and the heavier block be d and d' respectively, then by the kinematic eqn:
[tex]v^{2} = 2ad[/tex] (2)
[tex]v'^{2} = 2ad'[/tex] (3)
Dividing eqn (2) by (3), we get:
[tex](\frac{v}{v'}^{2}) = \frac{d}{d'}[/tex]
[tex]\frac{d}{d'} = (-3)^{2}[/tex]
[tex]\frac{d}{d'} = 9[/tex]
