A block of mass 1 kg moving with a speed of 4 m s - 1 , collides with another block of mass 2 kg which is at rest. The lighter block comes to rest after collision. The loss in the K . E . of the system will be

Question

A block of mass 1 kg moving with a speed of 4 m s - 1 , collides with another block of mass 2 kg which is at rest. The lighter block comes to rest after collision. The loss in the K . E . of the system will be, 8 J, 4 × 10 - 7 J, 4 J, 0 J

MARKS App · Accepted Answer

Let v 1 be the initial velocity of 1 kg block and v 2 be the final velocity of 2 kg block. From law of conservation of momentum, we get 1 v 1 = 2 ( v 2 ) ⇒ v 2 = v 1 2 ⇒ v 2 = 4 2 = 2 m s - 1 ∵ v 1 = 4 m s - 1 Δ K i n e t i c e n e r g y = 1 2 m 1 v 1 2 - 1 2 m 2 v 2 2 = 1 2 1 × 4 2 - 2 × 2 2 = 1 2 × 8 = 4 J

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