# Show that Momentum is conserved in a reference frame

1. Jun 7, 2015

### Tonia

1. The problem statement, all variables and given/known data
Show that momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

2. Relevant equations
Mass M for 2000 kg car, Mass m for 1500 kg car. To the observer on the ground: mv + 0 = (M + m) v v = Mv/M + m
The two cars are one mass after the collision (M + m), with velocity v:
v = Mv/M + m
After the collision, M + m is moving at a speed of v - v
M (V -v) - mv = (M + m) (v - v)
MV - Mv -mv + Mv + mv = (M + m) v v = Mv/(M + m)

3. The attempt at a solution
To the observer on the ground: mv + 0 = (M + m) v v = Mv/M + m
M (V -v) - mv = (M + m) (v - v)
MV - Mv -mv + Mv + mv = (M + m) v v` = Mv/(M + m)

Last edited by a moderator: May 7, 2017
2. Jun 7, 2015

### unscientific

Let's assume that the 2000kg car is moving towards the right initially. In the reference frame 10m/s to the right, how fast is each car moving before collision? After collision, they move off as one with a certain speed. What is that speed?

Last edited by a moderator: May 7, 2017
3. Jun 7, 2015

### Noctisdark

Hi there, to prove that momentum is conserved, show that Mv1 + mv2 = Mv'1 + mv'2, in the moving frame of reference whose direction is the velocity of the car M we don't have to worry about dealing with vectors, let it all be scalar where (+) is the direction of M, what are the new speeds of cars M and m ? good luck !!

4. Jun 7, 2015

### Tonia

Note: I wrote 2.00 kg car when I meant a 2000 kg car.
Before the collision, the larger car is moving at 20.0 m/s, and the car at rest is moving at 0.0 m/s. The speed they move (together) is: Mv/M+m = (2000 kg times 20.0 m/s) divided by (2000 kg + 1500 kg
= 40000 divided by 3500
= 11.4285714
= 11.43 m/s of cars together after collision.

5. Jun 7, 2015

### SammyS

Staff Emeritus
Be careful to use enough parentheses to say what you mean: Mv/(M+m) rather than Mv/M+m .

What you have so far is the result in a reference frame stationary with respect to Earth.

Take that result and calculate the momentum of each vehicle in the moving reference frame, before & after the collision. Is momentum conserved ?

6. Jun 7, 2015

### Tonia

I am not sure how to.

7. Jun 7, 2015

### SammyS

Staff Emeritus
The 2000 kg car travels at 20 m/s and the reference frame travels in the same direction at 10 m/s.

What is the velocity of the 2000 kg car with respect to the reference frame?

8. Jun 7, 2015

### Tonia

10 m/s?? Because 20 m/s - 10 m/s = 10 m/s?

9. Jun 7, 2015

### SammyS

Staff Emeritus
Right.

What is the velocity of the 1500 kg car with respect to the moving reference frame?

10. Jun 7, 2015

### Tonia

0 m/s?

11. Jun 7, 2015

### SammyS

Staff Emeritus
No.

If the 1500 kg car had the same velocity as the moving reference frame, the velocity of the 1500 kg car with respect to the moving reference frame would be 0 .

What is the velocity of the 1500 kg car with respect to the moving reference frame?

12. Jun 7, 2015

### Tonia

10m/s?

13. Jun 7, 2015

### SammyS

Staff Emeritus
In what direction?

If the 2000kg car and the reference frame move to the right with respect to Earth, that make their velocities both positive.

That also makes the 1500 kg car move to the left with respect to the moving reference frame. Correct?​

14. Jun 7, 2015

### Tonia

10 m/s west(left)

15. Jun 7, 2015

### SammyS

Staff Emeritus
Correct.

So, we would consider that to be -10 m/s .

What is the velocity of the two cars after the collision (in the moving frame) ?

16. Jun 7, 2015

### Tonia

11.43 m/s

17. Jun 7, 2015

### SammyS

Staff Emeritus
No. That's relative to Earth.

What is that relative to the moving reference frame?

18. Jun 7, 2015

### Tonia

10 m/s? I'm sorry I'm just having difficulty understanding it.

19. Jun 7, 2015

### SammyS

Staff Emeritus
After the collision, the two cars have a velocity (to the right, which is positive) of 11.43m/s relative to Earth.

The moving reference frame has a velocity of 10 m/s (to the right also).

What is the velocity of the two cars after the collision (relative to the moving reference frame) ?

20. Jun 7, 2015

### Tonia

11.43 m/s minus 10 m/s?