Grade 12 Physics Momentum

[Panphobia]

So I have my Unit 2 physics test tomorrow, and I was going through the review and I could not figure out how to do this question

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A 44 g bullet strikes a 1.54 kg block of wood placed on a horizontal surface just in front of the gun. The coefficient of kinetic friction is 0.28 and the impact moves the block a distance of 18.0 m before coming to rest. a) What was the normal force on the block? b)What was the force of friction on the block? c)What was the velocity of the block immediately after impact? e)What was the initial velocity of the bullet?

Now for a) I think the normal force would be mg, for b) I think the force of friction would be 0.28 * 18, I am not completely sure about that, and I do not know what to do for c and e, but I do now that kinetic energy is not conserved so we only use momentum.

[Insectoid]

A) That's correct.
B) Fk = uk * N. What you have is uk * d, which as far as I know isn't anything useful.
C) The only force acting on the block during deceleration is friction, and we know the mass. Since F = ma, we can calculate acceleration. Given acceleration, vf, and distance, we can calculate v0.
D) What? There's no D?
E)You know the reaction velocity and mass of the block, and the mass of the bullet. Given P=mv and P is conserved, you can calculate the initial velocity of the bullet.

Bear in mind I haven't done this in a while (Actually, I'm lying. I'm taking a college physics class right now, but it's way, way easier than my university or even high school physics were) so I might be forgetting something.

[DemonWasp]

a) The normal force is Fn = mg; presumably you would use the mass of the block plus bullet (because that's what's sliding).
b) The force of friction is Ff = umg = uFn (which you calculated in part (a))
c) Start by using your kinematics equations to turn the (friction force, mass) into (acceleration); then use (acceleration) and (distance) and (final velocity = 0) to get (initial velocity).
e) Use conservation of momentum.

[Insectoid]

Heh, I beat ya, DemonWasp.

Quote:

presumably you would use the mass of the block plus bullet (because that's what's sliding).

Hmm, I didn't think of that. I assumed the bullet bounced off or for whatever reason was not embedded in the block.

[Panphobia]

Oh crap on the review sheet there was no d, I just copied the question straight off, sorry. But thank you for your help, it cleared it up. Also, lets say if you had a spring how would you change the conservation of momentum m1v1 + m2v2 =..... to accommodate for the springs elastic potential energy 1/2kx^2, one example is

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A 3 kg mass and a 5 kg mass are resting on a frictionless surface and compressing a spring between the two masses. When the masses are released the 5.0 kg mass acquires a speed of 2.0 m/s. a) Find the speed of the 3 kg mass. b) How much energy was stored in the spring?

So would v1 and v2 be 0 and then would the equation be 0 = m1v1' + m2v2' + (would there be something here?), and to find the energy that was stored in the spring I do not know in the slightest.

[crossley7]

keep in mind it is conservation of momentum. Energy does not affect momentum so the elastic potential energy has no affect. you end up with 0=m1v1+m2v2 and solve for the 1 unknown. the elastic potential would be used if you were doing this using conservation of energy but clearly that is not the case here.