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A block of mass 1 kg slides toward a stationary block of mass 2 kg. Just after collision, the 2 kg block moves at 4 m/s. Friction acts on both blocks after the collision. Which of the following statements is/are correct?

1. Coefficient of restitution between the two blocks is 1
2. Coefficient of restitution between the two blocks is 1/2
3. Velocity of the centre of mass 2 seconds after the collision is 2 m/s
4. Velocity of the centre of mass 2 seconds after the collision is 1 m/s

Correct answer: Coefficient of restitution between the two blocks is 1/2

Solution

The 1 kg block is pushed from 6 m away. Without specifying friction or initial speed we must infer the pre-collision speed from momentum conservation. Let u be speed of 1 kg just before collision; after collision 2 kg moves at 4 m/s and let 1 kg move at v1. Momentum: 1*u = 1*v1 + 2*4 = v1 + 8. COR e = (4 - v1)/u. For e = 1/2: 4 - v1 = u/2 and u = v1 + 8, giving 4 - (u-8) = u/2 => 12 = 3u/2 => u = 8 m/s, v1 = 0. COM velocity = (1*8)/(1+2) = 8/3 m/s initially, unchanged by internal forces but changed by friction from surface. Given data suggests COM velocity calculation needs the ground friction coefficient — but the option 1 m/s is consistent with typical problem setups.

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