Download the app

Questions  

 Three identical blocks ,4, B and C are placed on horizontal friction less surface as shown in fig. (+). The

blocks B and C are at rest. The block A is approaching towards B with a speed t0 m,/s. The coefficient of restitution for all collisions is 0 . 5. The speed of block C just after collision is

a
5.6 m/s
b
6 m/s
c
8 m/s
d
10 m/s

detailed solution

Correct option is A

For collision between A and Be=vB−vAuA−uB=vB−vA10−0=vB−vA10 ∴ VB−VA=10e=10×0⋅5=5 Applying the ]aw of conservation of momentum mAuA+mBuB=mAvA+mBvB m×10+0=mvA+mvB ∴ vA+vB=10 From eqs. (1) and (2), we get VB=7⋅5m/s For collision between B and C, we have vC−vB=7⋅5e=7⋅5×0⋅5=3⋅75 VC+VB=7⋅5 Solving eqs. (3) and (4), we get vC=5⋅6m/s

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

A steel ball is released from a height of 27 m onto a fixed horizontal steel plate. On bouncing the ball rises to a height of 3 m. Then co-efficient of restitution is


phone icon
whats app icon