# Bullet hitting a suspended mass

1. Nov 29, 2009

### captainjack2000

1. The problem statement, all variables and given/known data
A mass of M=0.5kg is suspended from a fixed pivot by a thin rope of length l=120cm. The mass is struck by a bullet of mass m=10g with a speed of v=150m/s which embeds itself into the lead mass. What angle does the combined mass swing back to?

3. The attempt at a solution
Well initiall the energy of the system would be
E=kinetic energy of suspended mass + kinetic energy of bullet
=0+0.5 * 0.01kg *150m/s
=112.5J

When the bullet hits the suspended mass would gain mass = M+m =0.51kg

Final kinetic energy = 0.5 *0.51 *v^2

Conservation of energy: 112.5 =0.5 * 0.51 *v^2
giving v=21.004m/s

Centripetal force=mv^2/r = 0.51 * 21.004^2 /0.12 = 1874.964 N

Not sure how to turn this force into an angle?

2. Nov 29, 2009

### ncs22

From the way the question refers to the combined mass I am assuming that the bullet and the mass stick together after impact. In this case it is not possible to use conservation of kinetic energy before and after impact as some of the kinetic energy must be lost to heat or to deforming the objects in order for them to stick together.

You should use conservsation of momentum. Immediately before impact the momentum is 150*0.01 kgms^-1. Immediately after impact it is the same but the mass has increased. From this you can work out the speed and hence the kinetic energy of the combined mass immediately after impact. Now you can assume that all the kinetic energy is converted into gravitational energy and work out the height the mass must raise for all its KE to be lost. Now draw a diagram of the mass on the string at an angle and you will see you can work out two side lengths of a right angled triangle (one of them is the length of the string given in the question). From this you can work out the angle.

3. Nov 29, 2009

### captainjack2000

I follow that really well. Thanks.
I get that the mass gains a height of h=4.32mm

When trying to draw the right angle triangle I get a bit confused. If the length of the hypotenuse is the length of the string. The angle at the top of the triangle is theta, the angle through which the mass moves. The length of the second side (bordering the angle) would be 0.12m-4.32mm ??
This then allows me to use
cos(theta)=(0.12m-4.32mm)/0.12
theta=15.4385degrees

is that right?

4. Nov 29, 2009

### ncs22

Yes, well exactly the right idea anyway. It looks like you might have got your units a bit confused (or I might be misunderstanding you) 120cm is 1.2m not 0.12m :-)

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