# I'm stuck on momentum problems

1. Oct 13, 2006

### hbomb

A moving truck strikes a parked car. During the collision the cars stick together and slide to a stop together. The moving truck has a total mass of 2747 kg (including driver), and the parked car has a total mass of 1068.0 kg. If the cars slide 67.5 m before coming to rest, how fast was the truck going? The coefficient of sliding friction between the tires and the road is 0.400.

I know I need to use conservation of momentum somehow. But since I'm not given any of the velocities, how do I go about finding the initial velocity of the truck. Is there another equation out there that I have use?

2. Oct 13, 2006

### Staff: Mentor

First find the velocity of truck + car just after the collision. Calculate the acceleration (using Newton's 2nd law) and apply kinematics to calculate the speed.

3. Oct 13, 2006

### hbomb

That's what I've been stuck on, these queations are from a chapter on momentum and impulse. I'm suppose to use the conservation of momentum somehow.

4. Oct 13, 2006

### Locrian

Conservation of momentum in this case will read

$$m_1*v_1 + m_2*v_2 = m_1*v'_1+m_2*v'_2$$

Where the primeds denote the final velocities. If they stick together you know the two final velocities are equal. You also know how fast the car was initially going, so there are only two variables - how fast they end up going, and how fast the truck was originally going.

To figure out how fast they were originally going, use the work-energy theorem. You know the amount of work friciton did to stop them. Use that to find out how much kinetic energy the two had after the collision, and then you will know their final velocity.

These types of problems are common in the chapter you are working on - they combine a conservation of momentum problem with an energy problem.

5. Oct 13, 2006

### hbomb

Yeah, the only thing that I understood was the conservation of momentum. This problem is still unclear to me.

work-energy theorem is k=k(0)+W
and W=Fx
But with the coefficient of friction
UfW=Fx
so work-energy theorem becomes k=k(0)+Fx/Uf
so now what? this is as far as i have gotten

Last edited: Oct 13, 2006
6. Oct 13, 2006

### Locrian

I don't think you have the right formula for work due to friction, though I may just not understand your notation.

$$W_f = uNx$$

where N is the normal force and x is the distance traveled. You have enough information from what's given in the problem to solve that. From there you can solve for the velocity of the combined vehicles.

Let me know if it is still unclear.

7. Oct 14, 2006

### hbomb

Yea, I figured it out know. Thanks. I should of waited till the morning to work on this one, I was up late solving others problems. I do have another momentum-like problem that I have no idea how to get the answer.

Suppose a particle has an initial velocity of -1.51 km/s (moving in negative x-direction, to the left), and that the left wall moves with a velocity of 0.71 km to the right, and that the right wall moves with a velocity of 2.11 km to the left. What is the velocity of our particle after 8 collisions with the left wall and 7 collisions with the right wall?

Hint:You need to understand elastic collisions in one dimension in the limit that one of the two collision partners is practically infinitely heavier than the other one. In this limit, you need to look at the formula for the velocity of the lighter collision partner after the collision. Once you have solved this for one collision of our particle with a wall, it is easy to extend it to 7 or 8 collisions.

Last edited: Oct 14, 2006