Three identical blocks A , B and C are placed on a horizontal frictionless surface. The blocks B and C are…

Three identical blocks A, B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m s-1. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after the collision is


  1. 5.6 m s-1
  2. 6 m s-1
  3. 8 m s-1
  4. 10 m s-1

Solution

For collision between blocks A and B,

e=vB-vAuA-uB=vB-vA10-0=vB-vA10

∴vB-vA=10e=10×0.5=5      ….(i)

From principle of momentum conservation,

mAuA+mBuB=mAvA+mBvB

m×10+0=mvA+mvB

∴  vA+vB=10                     ….(ii)

Adding eqs. (i) and (ii), we get

vB=7.5 m s-1                       …(iii)

Similarly for collision between B and C,

vC-vB=7.5e=7.5×0.5=3.75

∴  vC-vB=3.75 m s-1 ...(iv)

Adding Eqs. (iii) and (iv) we get

2vC=11.25

∴  vC=11.252=5.6 m s-1

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