# Momentum in different referance frames

Murilo T
Homework Statement:
For some odd reason, you decide to throw baseballs at a car of mass ##M##, which is free to move frictionlessly on the ground. You throw the balls at the back of the car at speed ##u##, and at a mass rate of ##\sigma## (assume the rate is continuous, for simplicity). If the car starts at rest, find its speed and position as a function of time, assuming that the balls bounce elastically directly backwards off the back window.
Relevant Equations:
## v_{i} - V_{i} = - (v_{f} - V_{f})##
I aready got the solution for this exercise. However, the solution used the referance frame from the car:
Let the speed of the car be ##v(t)##. Consider the collision of a ball of mass ##dm## with the car. In the instantaneous rest frame of the car, the speed of the ball is ##u - v##. In this frame, the ball reverses velocity when it bounces, so its change in momentum is ##-2( u - v )dm##. This is also the change in momentum in the lab frame, because the two frames are related by a given speed at any instant. Therefore, in the lab frame the car gains a momentum of ## 2( u - v )dm## from each ball that hits it. The rate of change in momentum of the car is thus:
##\frac {dp} {dt} = 2\sigma'(u - v)##
where ##\sigma' = \frac {dm} {dt}## is the rate at which mass hits the car. ##\sigma'## is related to the given ##\sigma## by ##\sigma'= \frac {\sigma( u - v)} {u}##. We therefore have ##M \frac {dv} {dt} = \frac {2( u - v)^2\sigma} {u}##.
What I'm trying to understand is the line:
This is also the change in momentum in the lab frame, because the two frames are related by a given speed at any instant.
Because before reading the solution, I was trying to solve it using the lab frame.
So this is my work so far:

Using conservation of momentum and kinetic energy

##dmu + Mv = M(v +dv) - dmu##

##\frac {dmu^2} {2} + \frac {Mv^2} {2} = \frac {M(v + dv)^2} {2} + \frac {dmu^2} {2}##

this gives ##dv = 2( u - v )##. It is also possible to use the principle that the relative velocity of the two objects after a collision is the negative of the relative velocity before the collision: ##v_{i} - V_{i} = -(v_{f} - V_{f})##

But when I try to find the rate of change in momentum of the car:

##\frac {dp} {dt} = \frac {Mdv} {dt} = \frac {2M(u - v)} {dt}##

The ##\sigma'## and the ##\sigma## doesn't even show up.

I also tried to solve it finding the rate of change in momentum of the balls, and then using the negative of it as the rate of change in momentum of the car, but it gives:
##\frac {dp_{balls}} {dt} = \frac {dm(2u)} {dt} = 2u\sigma' = 2(u - v)\sigma##

##\frac {dp_{balls}} {dt} = \frac {-dp_{car}} {dt}##

So the rate of change in momentum of the car is:
##-2(u - v)\sigma##

Is it possible to solve it using the lab frame? And if so, what mistake am I making?

Last edited:

## Answers and Replies

Science Advisor
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Using conservation of momentum and kinetic energy

##dmu + Mv = M(V +dv) - dmu##
Perhaps think about this.

Murilo T
Perhaps think about this.
Sorry! It should be ##dmu + Mv = M(v +dv) -dmu##
I wrote the V instead of v

Science Advisor
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Sorry! It should be ##dmu + Mv = M(v +dv) -dmu##
I wrote the V instead of v
The capitalisation of ##v## was not the problem!

Murilo T
The capitalisation of ##v## was not the problem!
Since in the first collision the car has zero speed, ##v = 0## and the expression becomes ##dmu = Mdv - dmu##.
But the mass isn't really an infinitesimal mass and the velocity isn't really an infinitesimal velocity. So I think that it could be rewrited to: ##m_{i}u + Mv_{i-1} = Mv_{i} - m_{i}u##, and ##v_{0} = 0##.
But I can't see how it can be solved for ##m_{i}##, ##v_{i-1}## and ##v_{i}##, because we don't know how many balls there are.

Last edited:
Science Advisor
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Since in the first collision the car has zero speed, ##v = 0## and the expression becomes ##dmu = Mdv - dmu##.
But the mass isn't really an infinitesimal mass and the velocity isn't really an infinitesimal velocity. So I think that it could be rewrited to: ##m_{i}u + Mv_{i-1} = Mv_{i} - m_{i}u##, and ##v_{0} = 0##.
But I can't see how it can be solved for ##m_{i}##, ##v_{i-1}## and ##v_{i}##, because we don't know how many balls there are.
The ball doesn't rebound at the same speed in any reference frame. You have a higher separation speed after the collision than before, which is incompatible with conservation of energy.

Murilo T
Murilo T
The ball doesn't rebound at the same speed in any reference frame. You have a higher separation speed after the collision than before, which is incompatible with conservation of energy.
OMG, it looks like I was blind before!
Indeed, my conservation of energy: ##\frac {dmu^2} {2} + \frac {Mv^2} {2} = \frac {M(v + dv)^2} {2} + \frac {dmu^2} {2}##

gives ## \frac {Mv^2} {2} = \frac {M(v + dv)^2} {2}##!

The initial velocity of the ball and its final velocity are different:

##dmu + Mv = M(v + dv) - dmu'## (1.1)
##u - v = -(u' -(v + dv) )## (1.2)

Isolating ##u'## in 1.2: ##u' = u -2v - dv##
Plugging this result in 1.1 and solving for ##dv##: ##dv = \frac {2dm(u - v)} {M}##

Now using ##\frac {dp} {dt'} = M\frac {dv} {dt'}## and substituting dv for the result above: ##\frac {dp} {dt'} = 2\sigma'(u - v)##
##\sigma' = \frac {\sigma(u - v)} {u}##
So ##\frac {dp} {dt'} = M \frac {dv} {dt'} = \frac {2( u - v)^2\sigma} {u}##.

Thank you very much!

PeroK