Find x(t) for a car being pelted by rocks (1-D)

[Phantoful]

Homework Statement

Homework Equations

p=mv
K=(1/2)mv²
Integration and Derivation Methods
J=mΔv

The Attempt at a Solution

So far I set P_before=(Δm*u)+(MV_car), with V_car depending on the time that has passed (I'm not sure how I should find the function for V_car(t). For P_after I included the elastic collision of the rocks, = (V_car - u)Δm + MV_car. However, I don't know where to go from there, and I'm not sure if my momentum equations are correct. How would I find V_car(t)? Or is there a way to do this using Kinetic Energy equations?

 

[BvU]

Check your list of symbols. It isn't clear what you mean with $\Delta$m (small m). Some of the symbols in your problem statement don't occur in your relevant equations, etc.

If the continuous stream is too abstract, start an attempt with a single stone (mass c) bouncing off at t=0, then another at t=1... etc.

 

[PeroK]

This is not an easy problem, so you are going to have to do a lot of thinking! First, I would suggest working out roughly what happens (as the car speeds up):

1) I would look at the initial collision with one rock hitting the car, which would be at rest.
2) I would the think about a rock hitting the car when it is moving at some speed. Hint: can you reduce this problem to the same as 1)?
3) I would think about how many rocks hit the car once it is moving. Hint: something important in physics is relevant here.

Once you've thought it all through, then aim for your differential equation.

Note: you may be able to look up the answers to 1-3, from what you have already studied, or you could calculate them yourself.

 

[PumpkinCougar95]

Its a bit tricky but here is a simple way to think about it:

At a general time $t$ the car moves forward by $Vdt$ but the steady stream of particles move forward by $udt$. So How much mass does the car deflect off? Also, think about the initial and final velocities of that mass. Obviously, the initial velocities would be $u$, what would be the final?

 

[Phantoful]

the

This is not an easy problem, so you are going to have to do a lot of thinking! First, I would suggest working out roughly what happens (as the car speeds up):

1) I would look at the initial collision with one rock hitting the car, which would be at rest.
2) I would the think about a rock hitting the car when it is moving at some speed. Hint: can you reduce this problem to the same as 1)?
3) I would think about how many rocks hit the car once it is moving. Hint: something important in physics is relevant here.

Once you've thought it all through, then aim for your differential equation.

Note: you may be able to look up the answers to 1-3, from what you have already studied, or you could calculate them yourself.

1. So for the first rock, I got the equation m_rock*u = v_c*M + m_rock*f, with v_c being the speed of the car after getting hit, and f being the speed of the rock after hitting the car elastically. For the first rock, should I be assuming that f is equal to u? I'm not sure if it is, because once the rock collides the car will move, and then f would need to be less than u. (Actually should f be in reference to the frame of the car?)

2. For this one I'm not sure if initially my momentum equations should all be in respect to the car, or the person throwing the rocks. If I use the person, initially p = m_rock*u once again, or should I make it p = m_rock*(u-v_c), because the rock's speed is (u-v_c) with respect to the car's frame (correct?). I'm going to try the car frame, since it makes sense to me. And if it's all in the frame of the car, f, the speed the rock has from an elastic collision, would be something less than (u-v_c)

Should I use Kinetic energy to solve for everything? I'm not understanding what to do.

1. (1/2)(m_rock*u²) = E_init
E_final = (1/2)(M(v_c)²)+1/2(m_rock*f²)

And would you solve for f? Or am I going about it wrong?

 

[Phantoful]

Actually, the question says that the rock leaves with the same velocity it hits with, because it's so small, and conservation of momentum is basically ignored? I don't get how if the rock is moving at the same speed coming off, the car could be moving at all.

 

[PeroK]

1. So for the first rock, I got the equation m_rock*u = v_c*M + m_rock*f, with v_c being the speed of the car after getting hit, and f being the speed of the rock after hitting the car elastically. For the first rock, should I be assuming that f is equal to u? I'm not sure if it is, because once the rock collides the car will move, and then f would need to be less than u. (Actually should f be in reference to the frame of the car?)

2. For this one I'm not sure if initially my momentum equations should all be in respect to the car, or the person throwing the rocks. If I use the person, initially p = m_rock*u once again, or should I make it p = m_rock*(u-v_c), because the rock's speed is (u-v_c) with respect to the car's frame (correct?). I'm going to try the car frame, since it makes sense to me. And if it's all in the frame of the car, f, the speed the rock has from an elastic collision, would be something less than (u-v_c)

Should I use Kinetic energy to solve for everything? I'm not understanding what to do.

1. (1/2)(m_rock*u²) = E_init
E_final = (1/2)(M(v_c)²)+1/2(m_rock*f²)

And would you solve for f? Or am I going about it wrong?

I'm wondering whether this problem is a bit advanced. You should be able to calculate what happens in a simple collision with the car at rest. If that's difficult, then this problem is a whole lot harder.

In your method, you can use the KE equations, of course. But, you were given an alternative, simpler way to calculate the velocities in an elastic collision in the question: using separation velocities.

That said, you might want to take a look at what @PumpkinCougar95 said in post #4. That's definitely a simpler way to look at the whole problem.

 

[PeroK]

Actually, the question says that the rock leaves with the same velocity it hits with, because it's so small, and conservation of momentum is basically ignored? I don't get how if the rock is moving at the same speed coming off, the car could be moving at all.

That's in the reference frame of the car.

 

[jbriggs444]

Actually, the question says that the rock leaves with the same velocity it hits with

If you read the problem, it is fairly careful on this point. The impact velocity relative to the pre-impact car is equal to the egress velocity relative to the post-impact car.

The rock mass is small relative to the car, so one is invited to handwave over the distinction between pre-impact and post-impact when computing the rock's egress velocity. One could make that handwave rigorous when integrating instead of summing.