Paragraph: Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 ~m / s . The block undergoes a collision with stationary block B , which is connected to a non-deformed spring of spring constant 1000 Nm ⁻¹ . The coefficient of restitution between the blocks is 0.5 . After the collision, block B oscillates in SHM with a period of 0.2 s , and block A slides off t

Step-by-step solution

Immediately after the collision, suppose velocities of the blocks are v₁ and v₂ as shown 1 2 (velocity of approach) = velocity of separation. 5=v₂-v₁ (i) Using principal of conservation of momentum for the collision 2=0.2 v ₁+ v ₂ 10= v ₁+5 v ₂ (ii) On solving v ₂=2.5 ~m / s and v ₁=-2.5 ~m / s . Hence block A moves leftward after the collision with speed 2.5 ~m / s . And the block B moves towards right with speed 2.5 ~m / s . The maximum velocity of B=2.5= A A = v m k =2.5 1 1000 ~m =2.5 10 ~cm