# Find coefficient of kinetic friction between a bullet and a pendulum

1. Nov 11, 2012

### jreelawg

1. The problem statement, all variables and given/known data

A bullet collides and embeds itself into a block hanging from a rope. The block and embedded bullet swing out to an angle of 10 degrees.

If the bullet plows 2 cm into the block before stopping, what is the coefficient of kinetic friction between the block and the bullet?

bullet has mass of .1 kg
block on pendulum of length 1 m has mass of 5 kg

2. Relevant equations

momentum conservation, energy conservation, newtons laws

3. The attempt at a solution

I found the initial velocity of the bullet, 273.386 m/s^2, using momentum and energy conservation.

Thought maybe I could use, vf^2 = vi^2 + 2ax, which I guess tells me that the bullet after collision had an acceleration in the x direction of -1,868,500 m/s^2 with respect to the frame of reference of the pendulum+embedded bullet.

I thought maybe I would then use F=ma.

2. Nov 11, 2012

### Eango

I think your on the right track.
Would'nt it just be from where you left off:
Ffriction = ma
Ff = ma
UN = ma
Umg = ma
U = ma/mg
U = a/g
Also make sure to take the 10 degree into consideration and i think your good then.

3. Nov 11, 2012

### haruspex

Wrong units - should be m/s. And I get rather less than 273m/s. Please show your working to this point.

4. Nov 11, 2012

### jreelawg

I accidentally used .01 instead of .1 for mass of bullet in my calculation.

Here is what I have. I'm especially unsure about the validity of the last part.

v1 = velocity immediately after collision
v0 = initial velocity of bullet
mp = mass of pendulum = 5 kg
mb = mass of bullet = 0.1 kg
y = delta Y = 1 - cos(10 degrees)

momentum conservation:

mb*V0 = (mb + mp)*v1
v0 = (mb + mp)(v1)/mb

energy conservation of swing:

1/2(mb + mp)*v1^2 = (mp + mb)*yg
v1 = (2yg)^(1/2)

=> v0 = (mb + mp) (2yg)^(1/2) / mb
=((.1 + 5) (2*(1-cos(10))*9.8)^(1/2) )/0.1

= 27.8297 m/s

Acceleration of bullet after collision:

0 = 27.8297^2 + 2*a*(0.02)
a = -(27.8297^2) / 0.04)
= -19326.3 m/s^2

Force:

Fnet on bullet = Fk = -9.8 * 0.1 * Uk = -19326.3 * 0.1

Uk = 1972.07

My thoughts:

I don't think the last part is right. I would think that the main "normal" force affecting the friction would be the block squeezing the bullet as it embeds itself into the block. The coefficient of kinetic friction I would think would be a constant determined by the two materials. But this problem doesn't lend any information about either the force which would squeeze the bullet or specific materials.

Last edited: Nov 11, 2012
5. Nov 11, 2012

### jreelawg

Could this be solved using the work kinetic energy theorem?

6. Nov 11, 2012

### jreelawg

So kinetic energy is lost to thermal energy.

1/2(mb)V0^2 - 1/2(mb + mp)V1^2 = Eth ?
= 37.9634 J = fk (.02) = n * Uk
uk = (37.9634/.02)/n = 1898.17/n

So is n really just the (mb)g?

7. Nov 11, 2012

### haruspex

Agreed. (Though this is making the simplifying assumption that the bullet comes to rest so quickly within the block that most of the swing is with the bullet and block moving as one. Tricky question otherwise.)
Again, I agree. You can calculate the retardant force on the bullet, but that still doesn't tell you the coefficient of friction.

8. Nov 12, 2012

### jreelawg

Ok. Thanks haruspex. I guess I'll put this problem to rest for now.