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To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0126-kg bullet is fired straight up at a falling wooden block that has a mass of 2.76 kg. The bullet has a speed of 592 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t the tolerance is +/-2% the

Explanation / Answer

Solution:

Using the conservation of momentum

=> Initial momentum = final momentum

=> m* Vbu + M * (Vbl) = (m + M) V ...........(i)

Vbu = velocity of bullet = 592 m/s

Vbl = velocity of block

V = velocity of block + bullet

=> Now Vbl and V should be same as the block reach at the top in same time.

V = Vbl = g*t

Using equation (i)

=> 0.0126 * 592 - 2.76 * V = (0.0126 + 2.76) * V

=> 7.4592 = 5.5326 * V

=> g * t = 7.4592 / 5.5326

=> t = 0.138 sec (appr0x)

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