Chapter 07, Problem 24 To view an interactive solution to a problem that is simi

Question

Chapter 07, Problem 24

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.00720-kg bullet is fired straight up at a falling wooden block that has a mass of 1.97 kg. The bullet has a speed of 584 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.

Explanation / Answer

Given that -

m1 = 0.00720 kg
m2 = 1.97 kg
u1 = 584 m/s
u2 = 0m/s

Now -
at time t=t
v1 = u-gt = 584 - gt
v2 = -gt
after collision the objects combine
M = 1.97 + 0.00720 = 1.9772 kg
v = V
v=u+at
0=V-at
V=gt (t will be the same)

p1+p2=p1'+p2' {' means final i .e after collision}
m1v1+m2v2=(m1+m2)V

put the values -

0.00720*(584 - 9.8t) + 1.97*(9.8t) = (1.9772*gt)
=> 4.205 - 0.07056t + 19.306t = 19.38t

=> 4.205 = 0.1411t

=> t = 29.8 s.

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