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A 20 g bullet traveling at 600 m/s strikes a 4 kg block suspended by a string. If, after the impact, the block swings up to a height of 0.2 m, what is the speed of the bullet as it emerges from the block?

1. 200 m/s
2. 150 m/s
3. 400 m/s
4. 300 m/s

Correct answer: 200 m/s

Solution

Block speed after impact V = sqrt(2*10*0.2) = 2 m/s. Momentum: 0.02*600 = 0.02*v + 4*2, so 12 = 0.02v + 8, giving v = 200 m/s for the emerging bullet.

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