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A block A of mass 2 kg rests on a horizo...

A block A of mass 2 kg rests on a horizontal surface. Another block B of mass 1 kg moving at a speed of m/s when at a distance of 16 cm from A. collides elastically with A. The coefficeint of friction between the horizontal surface and earth of the blocks is 0.2. Then `(g = 10 m//s^(2))`

A

after collision block B rebounds

B

after collision block B comes to rest

C

final separation between the blocks is 30 cm

D

final separation between the blocks is 50 cm

Text Solution

Verified by Experts

The correct Answer is:
A, D

Vel of B, before colliding
`V^(2)=I^(2) -2(0.2xx10)(16)/(100)=1-64=36`
`V = 0.6 m//s`
After collision `V_(A)=((1+1)(0.6))/(3)=0.4 m//s`
`V_(B)=-0.2 m//s`
Let `S_(A)` be the disp. OF A, after collision
`S_(A) = (0.4^(2))/(2(2)) = (16)/(4) = 0.4 m`
Similarity `S_(B)=(0.2^(2))/(4)=(0.4)/(4)=0.1`
`:.` sepanation `=50cm`.
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