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Question

Two blocks of masses 5 kg and 2 kg are placed at rest on a frictionless surface and connected by a spring. An external hit gives a velocity of 21 m/s to the heavier block towards the lighter one. Velocities of both blocks (heavier and lighter one) in the centre of mass frame, just after the kick will be respectively:
(Consider direction of motion of heavier block as +ve direction)
  1. 6 m/s, 10 m/s
  2. 15 m/s, 15 m/s
  3. 6 m/s,15 m/s
  4. 3 m/s, 5 m/s


Solution

The correct option is C 6 m/s,15 m/s

vCM=m1v1+m2v2m1+m2
vCM=(5×21)+05+2=15 m/s
From the frame of system's COM, centre of mass of system will remain at rest (no external force)
Velocity of the heavier block in COM frame, just after the kick is
v1=v1vCM
v=(2115)=6 m/s
& Velocity of the lighter block in COM frame ,just after the kick is
v2=v2vCM
v2=(015)=15 m/s

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