A bullet of mass 0.1kg and speed v passes completely through a pendulum bob of m

Question

A bullet of mass 0.1kg and speed v passes completely through a pendulum bob of mass 2kg. the bullet emerges with a speed of v/2. the pendulum bob is suspended by a stiff rod of length 1.0m and negligible mass.

a. assuming that the velocity of the bullet was 200m/s when it entered the bob and 100m/s when it left, what is the velocity of the bob just after the bullet passes through it?

b. how high does the bob rise before coming to rest momentarily?

c. what is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Explanation / Answer

(a) Conservation of momentum says that the total momentum of the bullet and bob initially has to equal the total momentum after the bullet passes through the bob.

m1v1i + m2v2i = m1v1f + m2v2f

If m1 is the bullet and m2 is the bob, v2i = 0, since the bob is intially at rest. Plugging in the values given in the problem gives you:

(0.1 kg)(200 m/s) + 0 = (0.1 kg)(100 m/s) + (2 kg)v2f

v2f = 5 m/s

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(b) To find the height of the bob after the bullet leaves, you need to use conservation of mechanical energy:

Ki + Ui = Kf + Uf

Where K is the current kinetic energy:

K = (1/2)mv2

and U is the current gravitational potential energy:

U = mgh

Putting these into the energy conservation equation gives you:

(1/2)mvi2 + mghi = (1/2)mvf2 + mghf

Here, the bob's initial velocity is the velocity after the bullet leaves, which was found in part (a). Take the intial height to be 0, and the final velocity will be 0 at the bob's maximum height. Plugging in these values gives you:

(1/2)(2 kg)(5 m/s)2 + 0 = 0 + (2 kg)(9.8 m/s2)hf

hf = 1.276 m

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(c) For this part, you need to look at the centripetal acceleration of the bob. Centripetal acceleration is expressed as:

ac = v2/r

Here, r is the length of the pendulum. Plugging in what you know so far gives you:

ac = v2/(1 m)

v2 = ac

In order to find the minimum tangential velocity, you must know the minimum centripetal acceleration needed to overcome the affect of gravity on the bob. If you look at the forces acting on the bob at the very top of the loop, you have the centripetal force pointing straight up and the force of gravity pointing straight down. The sum of these forces has to be 0 since the bob is constrained to move in a circle of constant radius. This gives you the equaition:

mac - mg = 0

ac = g = 9.8 m/s2

So, this tells you that the centripetal acceleration has to be at least g to make it all the way around the loop. Plugging this into the previous equation gives you:

v2 = ac = 9.8 m/s2

v = 3.13 m/s

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