Momentum Problem

Pughball

Wow, I'm thoroughly stumped on 2 questions. I'd appreciate any help I can get because this is a distance ed course and there's only answers no solutions to the questions.

1) a 25g bullet strikes and is embedded in a 1.35kg block placed on a horizontal surface. Coefficient of kinetic frriction is 0.25 and the impact drives the block 9.5m before it comes to rest. What was the muzzle speed on the bullet?

I've figured out the normal force, and the force of friction 3.37N. The deceleration is 2.45m/s/s. Now I'm at an absolute stalemate because I don't have initial velocity or the time it travels in order to figure out the momentum after the collision. PLEASE help.

2) Two marble spheres of masses 30 and 20 grams are suspended from the ceiling by massless strings. The lighter sphere is pulled aside through an angle of 75 degrees and let go. IT swings, collides elastically with the other sphere at the bottom of the swing.
a)to what maximum angle will the heavier sphere swing?
b) to what maximum angle will the lighter sphere swing?

Thanks so very much.

topsquark

Forum Staff
1) a 25g bullet strikes and is embedded in a 1.35kg block placed on a horizontal surface. Coefficient of kinetic frriction is 0.25 and the impact drives the block 9.5m before it comes to rest. What was the muzzle speed on the bullet?

I've figured out the normal force, and the force of friction 3.37N. The deceleration is 2.45m/s/s. Now I'm at an absolute stalemate because I don't have initial velocity or the time it travels in order to figure out the momentum after the collision. PLEASE help.
You appear to have the friction force correct. (You did remember to factor in the mass of the bullet, right? Though this only modifies the answer slightly.) Now that you've found the acceleration, I'd recommend finding the initial speed of the block/bullet combination using
$$\displaystyle v^2 = v_0^2 + 2ad$$

$$\displaystyle 0 = v_0^2 - 2(2.45)(9.5) \implies v_0 = 6.82~m/s$$

So now you know the speed of the bullet/block combination just after the collision. Can you find the initial speed of the bullet from here using conservation of momentum?

-Dan

topsquark

Forum Staff
2) Two marble spheres of masses 30 and 20 grams are suspended from the ceiling by massless strings. The lighter sphere is pulled aside through an angle of 75 degrees and let go. IT swings, collides elastically with the other sphere at the bottom of the swing.
a)to what maximum angle will the heavier sphere swing?
b) to what maximum angle will the lighter sphere swing?
This problem employs both conservation of momentum and conservation of energy.

First we need to find out how fast the lighter marble is moving when it strikes the heavier marble. So let's define the zero point for the gravitational potential energy to be the level where the two marbles collide. So we know that the lighter marble starts with a height of
$$\displaystyle h_0 = L - L~cos(75)$$
where L is the length of the string. (Notice that we don't know what this is.)

So using conservation of energy on the lighter marble:
$$\displaystyle K_0 + P_0 = K + P$$

$$\displaystyle mgh_0 = \frac{1}{2}mv^2$$

$$\displaystyle v = \sqrt{\frac{2L(1 - cos(75))}{g}} \approx 0.3889 \sqrt{L}$$

Now we can use conservation of momentum and conservation of energy (the collision is elastic) to find the speeds of both marbles after the collision. I won't go through the whole derivation, but we start with
$$\displaystyle m_1v_{01} + m_2v_{02} = m_1v_1 + m_2v_2$$
and
$$\displaystyle \frac{1}{2}m_1v_{01}^2 + \frac{1}{2}m_2v_{02}^2 = \frac{1}{2}m_1v_{1}^2 + \frac{1}{2}m_2v_{2}^2$$

and we can derive that
$$\displaystyle v_1 = v_{01} \frac{m_1 - m_2}{m_1 + m_2}$$
and
$$\displaystyle v_2 = 2v_{01} \frac{m_1}{m_1 + m_2}$$
when $$\displaystyle v_{02} = 0$$

So I get that
$$\displaystyle v_1 = - 0.07778 \sqrt{L}$$
and
$$\displaystyle v_2 = 0.31112 \sqrt{L}$$
using 1 as the lighter marble and 2 as the heavier marble. Don't worry about that negative sign on the lighter marble's velocity: that merely means that the marble has rebounded.

I'll leave it to you to finish up: it's just the reverse of the first step. And don't worry, that L will cancel itself out. If you have any troubles finishing, just let me know.

-Dan

Pughball

Thanks so much, I knew that first question was easier than I was making it out to be. My brain was fried after 11 hours of physics though, I just wasn't using that equation which you so helpfully furnished. I'm going to work on both of them now, thanks again!