# Bullet Block Collision

1. Nov 1, 2005

### marialo

I'm having trouble with this physics problem:

A bullet traveling horizontally with a velocity of magnitude 400 m/s, is fired into wooden block with mass 0.800 kg initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 m/s. the block slides a distance of 45 cm along the surface from its initial position. A) what is the coefficient of kinetic friction between the block and surface? B) what is the decrease in kinetic energy of the bullet? C) what is the kinetic energy of the block at the instant after the bullet passes through it?

i started doing part A, and found that the velocity while the bullet is in the block is 1.99 m/s. i also assumed that the work done by friction was equal to the change in kinetic energy of the system. but when i used this information to solve for the coefficient of friction i got something way too high. the answers are supposed to be : A) 0.222, B)-291J, C)0.784 J, but i have no idea of how to set this problem up. Please Help!!!

2. Nov 1, 2005

### Staff: Mentor

What's the mass of the bullet?

3. Nov 1, 2005

### marialo

Sorry, i forgot to metion that the bullet is 4.0 g, or 0.004 kg

4. Nov 1, 2005

### Staff: Mentor

First thing to do: Find the speed of the block immediately after the bullet passes through it, before it starts its slide along the surface. (Hint: Apply conservation of momentum.)

5. Nov 1, 2005

### lightgrav

did you try to split up original KE of bullet
between bullet and block?
(1.99 m/s too high for block)

6. Nov 1, 2005

### marialo

i already did that. since momentum is conserved, the momentum of the bullet initially is equal to the sum of the momentums after the collision. i found that the velocity off the block after collision was 1.4 m/s.

7. Nov 1, 2005

### lightgrav

knowing the block's KE after bullet passes,
and Work done by friction (formula),
what's the Friction *Force*?

8. Nov 1, 2005

### marialo

The work done be friction is d*mg(mu), where mg(mu) is the force of friction. the block's KE is just 1/2mv^2, right?

9. Nov 1, 2005

### Staff: Mentor

Now that you have the speed of the block, use kinematics to find the acceleration and required friction force.

10. Nov 1, 2005

### lightgrav

yea, KE known, mu is straightforward.

Because LOCATIONS were known/wanted,
this was best solved using an Energy approach.
momentum (or F=ma) would've been best.

(Why is the Change of bullet KE NOT = KE block?)

Last edited: Nov 1, 2005
11. Nov 1, 2005

### marialo

Sorry, i'm still having trouble seeing how to go from KE to mu

12. Nov 1, 2005

### lightgrav

umm, the KE of the block is removed by the
(negative) Work that the friction Force does on the block.

You got the block's KE before sliding, right?
(Where does that KE go AS it SLIDES?)

13. Dec 7, 2008

### Insomniac18

Alright, so velocity of the block is 1.4 m/s as marialo stated and I also found.
so since KE is the only thing that changes, KE=Work of Friction.
Work of friction = (8mu)(.45)(-1) = -3.6mu
KE = (.8/2)(1.4)^2=.784
mu = .217777...
Why didn't I get .222 as the book states to be the answer? Did I do anything wrong or it's a negligible error?