Linear Momentum Pellet Gun Problem

[maniacp08]

Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. On one particular shot, his 1.1 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is 76 m/s and the cheese has a mass of 136 g, what is the coefficient of friction between the cheese and ice?

Relevant equations:
Momentum = Mass * Velocity

This would be an example of an inelastic collision where the two objects stick together.
The KE is not conserved.

I converted the mass of the pellet and the cheese to kg.
Mass of pellet = .0011kg
Mass of cheese = .136kg

Momentum of the pellet = .0011kg * 76m/s = .0836 kg m/s
The momentum will be conserved because there is no external forces.

This is as far as I got, can someone tell me how should I approach this next?
Thanks.

 

[Doc Al]

Since momentum is conserved, find the speed of the cheese+pellet immediately after the collision. Then apply some kinematics to find the acceleration due to friction.

 

[maniacp08]

So I have mass of the pellet * Vi = mass of pellet + mass of cheese * Vf
.0011kg * .76m/s = (.0011kg + .136kg) * Vf
Vf = .61m/s

The only kinematics equation that do not deal with time is
Vf^2 = Vi^2 + 2 * A * D
I converted 25cm to .25m to be consistent

(.61m/s)^2 = (.76m/s)^2 + 2 * A * (.25m)
I got A = -.41m/s

Is this right to have a negative acc.?

If this is all right, do I continue with Newtons 2nd law to solve for the coefficient of friction?
F = MA?

 

[Doc Al]

So I have mass of the pellet * Vi = mass of pellet + mass of cheese * Vf
.0011kg * .76m/s = (.0011kg + .136kg) * Vf
Vf = .61m/s

Your formula is correct, but redo your calculation being careful with the decimal point.

The only kinematics equation that do not deal with time is
Vf^2 = Vi^2 + 2 * A * D
I converted 25cm to .25m to be consistent

Good.

(.61m/s)^2 = (.76m/s)^2 + 2 * A * (.25m)
I got A = -.41m/s

Several problems: The speed immediately after the collision becomes your initial velocity for this part of the problem. After the collision, the .76 m/s is no longer relevant. What's the final speed of the block of cheese?

Is this right to have a negative acc.?

Sure. That just means the cheese was slowing down.

If this is all right, do I continue with Newtons 2nd law to solve for the coefficient of friction?
F = MA?

Yes.

 

[maniacp08]

Yeah, I put it as .76m/s instead of 76m/s =[ Clumsy me.

The new final velocity would be 0 since it eventually stops.
0 = (.61m/s)^2 + 2 * A * (.25m)
I have A = -.74m/s

2nd part I have the forces in the Y direction:
Fn = MG where M = mass of cheese + mass of pellet
Fn = (.1371)kg*(9.81)m/s = 1.3J

The forces in the X direction:
There is just the force of friction correct?
so is
-Uk(Fn) = M * A
-Uk(1.3J) = .1371kg * (-.74m/s)
Uk = .08

Am I correct?

 

[Doc Al]

My only suggestion is not to round off intermediate calculations--wait until the last step. Other than that, perfect!

 

[maniacp08]

Thanks Doc for your help =].