a bullet(.010 kg) is fired straight up at a falling blockof wood (2.0 kg); bulle

Question

a bullet(.010 kg) is fired straight up at a falling blockof wood (2.0 kg); bullet has a speed of 750 m/s when it hits theblock. the block had been dropped from rest on top of abuilding. after the collision, the block and bullet rises upand comes to a halt at exactly where it had been dropped, thenbegins to fall again. How long had the block been fallingbefore the collision? have difficultly setting up this problem a bullet(.010 kg) is fired straight up at a falling blockof wood (2.0 kg); bullet has a speed of 750 m/s when it hits theblock. the block had been dropped from rest on top of abuilding. after the collision, the block and bullet rises upand comes to a halt at exactly where it had been dropped, thenbegins to fall again. How long had the block been fallingbefore the collision? have difficultly setting up this problem

Explanation / Answer

The mass of the bullet is m = 0.010 kg The mass of the falling block of wood is M = 2.0 kg The speed of the bullet is u = 750 m/s when it hits theblock According to the law of conservation of momentum,we have m * u + M * u1= (m + M) * V---------------(1) Here,u1= 0 m/s Substituting the values in equation (1),we get 0.010 * 750 + 2.0 * 0 = (0.010 + 2.0) * V or V = (0.010 * 750/(0.010 + 2.0)) or V = 3.73 m/s Let the time for which the block had been falling before thecollision be t.Therefore,we get V = u1 + gt Here,g = -9.8 m/s2 or t = (V - u1/g) or t = (3.73 - 0/-9.8) = 0.38 s or t = (V - u1/g) or t = (3.73 - 0/-9.8) = 0.38 s

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