A 7.00-g bullet, moving horizontally with an initial speed v0, embeds itself in

Question

A 7.00-g bullet, moving horizontally with an initial speed v0, embeds itself in a 1.20-kg pendulum bob that is initially at rest. The length of the pendulum is L = 0.710m . After the collision, the pendulum swings to one side and comes to rest when it has gained a vertical height of 14.0cm .

Part A

Is the kinetic energy of the bullet-bob system immediately after the collision greater than, less than, or the same as the kinetic energy of the system just before the collision?

Part B

Find the initial speed of the bullet. (m/s)

Part C

How long does it take for the bullet-bob system to come to rest for the first time? (s)

Explanation / Answer

Momentum's conservation :
m Vo = (m + M) Vf.
Energy's conservation , after collision:
(1/2)(M + m) Vf^2 = (M + m) g h
Vf = sqr( 2 g h) = sqr(2*9.8*0.14) = 1.6565 m/s
Vo = (M + m) Vf/m = 285.6279 m/s.
The period of the pendulum is
T = 2 pi sqr(L/g) = 1.69 s
Then t = T/4 = 0.42258 s
The energy lost in the collision is:
Elost = (1/2) m Vo^2 - (1/2) (M + m) Vf^2
Elost = |Work of friction|

The kinetic energy of the bullet-bob system immediately after the collision is less than the kinetic energy of the system just before the collision

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