Solving Baseballs Colliding with a Car: Find Speed & Position

[Nate Stevens]

Homework Statement

You throw baseballs at a mass rate of σ kg/s (assume a continuous rate) with speed u at the back of a car. They collide elastically with the car. The car starts at rest and has mass M. It moves on the ground with no friction. Find its speed and position as a function of time.

Homework Equations

F=ma=mdv/dt=dp/dt

The Attempt at a Solution

To start, I thought I should consider a single collision between one baseball and the car after it has started moving.
I let the thrown baseball have a mass of dm. Then when it hits the car, the car will only gain an infinitesimal amount of speed.
So the initial velocities before the collision are
baseball -> u
car -> v
And the final velocities after the collision are
baseball -> ??
car -> v + dv
If I use the fact that in elastic collisions the relative velocities of two objects (before and after the collision) are reversed, then
u-v = - ( ??-( v+dv ))
solving for ??
?? = 2v+dv-u
Now that we have the baseball's initial and final velocity we can get it's change in momentum
Δp = p_f - p_i = (2v+dv-u)×dm - (u)×dm = dm×dv + (2v-2u)×dm

I know from here I can get the force on the ball by differentiating the above momentum with respect to t, but before that I need to figure out what to do with the dm×dv. Is it okay say that this will go to zero since it is two very small values multiplied together?

If that is the case, are the final steps just rewriting dm/dt as σ, plugging the momentum into
mdv/dt=dp/dt, and separating variables twice? Somehow I don't think so, since σ is just the mass rate of baseballs being thrown and NOT the mass rate of baseballs hitting the car.

Even if it is just answering the dm*dv question, any help would really be appreciated.

 

[doggydan42]

Homework Statement

You throw baseballs at a mass rate of σ kg/s (assume a continuous rate) with speed u at the back of a car. They collide elastically with the car. The car starts at rest and has mass M. It moves on the ground with no friction. Find its speed and position as a function of time.

Homework Equations

F=ma=mdv/dt=dp/dt

The Attempt at a Solution

To start, I thought I should consider a single collision between one baseball and the car after it has started moving.
I let the thrown baseball have a mass of dm. Then when it hits the car, the car will only gain an infinitesimal amount of speed.
So the initial velocities before the collision are
baseball -> u
car -> v
And the final velocities after the collision are
baseball -> ??
car -> v + dv
If I use the fact that in elastic collisions the relative velocities of two objects (before and after the collision) are reversed, then
u-v = - ( ??-( v+dv ))
solving for ??
?? = 2v+dv-u
Now that we have the baseball's initial and final velocity we can get it's change in momentum
Δp = p_f - p_i = (2v+dv-u)×dm - (u)×dm = dm×dv + (2v-2u)×dm

I know from here I can get the force on the ball by differentiating the above momentum with respect to t, but before that I need to figure out what to do with the dm×dv. Is it okay say that this will go to zero since it is two very small values multiplied together?

If that is the case, are the final steps just rewriting dm/dt as σ, plugging the momentum into
mdv/dt=dp/dt, and separating variables twice? Somehow I don't think so, since σ is just the mass rate of baseballs being thrown and NOT the mass rate of baseballs hitting the car.

Even if it is just answering the dm*dv question, any help would really be appreciated.

The relative velocities should not be reversed. Use conservation of momentum as it applies to elastic collisions. The car and the baseball have different masses, so if the baseball moves very quickly, it might give the car a very small velocity since the mass of the car is much larger than the mass of the baseball.

 

[haruspex]

Is it okay say that this will go to zero since it is two very small values multiplied together?

Yes.

σ is just the mass rate of baseballs being thrown and NOT the mass rate of baseballs hitting the car.

Well spotted. How can you derive the one from the other? Hint: it is like a Doppler effect.

 

[haruspex]

The relative velocities should not be reversed.

Nate's equation is correct. This is a result of conservation of momentum in elastic collisions.

 

[Nate Stevens]

Yes.

Well spotted. How can you derive the one from the other? Hint: it is like a Doppler effect.

I have started drawing the case when two baseballs are thrown, and

from here I would imagine that you solve for the frequency of baseballs hitting the car. I just don't know how I can relate the frequency of balls thrown (inverse Δt) and the frequency of balls hitting the car (inverse Δt') to the mass rate of balls being thrown (σ) and the mass rate of balls hitting the car (σ' let's say).

Thanks for taking time to help me dan and haruspex!

 

[PeroK]

from here I would imagine that you solve for the frequency of baseballs hitting the car. I just don't know how I can relate the frequency of balls thrown (inverse Δt) and the frequency of balls hitting the car (inverse Δt') to the mass rate of balls being thrown (σ) and the mass rate of balls hitting the car (σ' let's say).

Thanks for taking time to help me dan and haruspex!

You seem to be nearly there. If you imagine a small mass $\Delta m$ is fired in a time $\Delta t$, then the same mass hits the car in $\Delta t'$. That's what you've calculated.

This should allow you to relate $\sigma'$ to $\sigma$.