# Speed of car related to momentum

1. Feb 7, 2010

### songoku

1. The problem statement, all variables and given/known data
A 1000 kg car with 4 people on it, each 75 kg, is at rest. The people jump off the car with speed 2 ms-1 relative to the car. Find the speed of the car if the people jump off :

a. simultaneously
b. one by one

2. Relevant equations
conservation of momentum

3. The attempt at a solution

a.
m1u1 + m2u2 = m1V1 + m2V2

0 = 300 * 2 - 1000 * v2

v2 = 0.6 ms-1

b.
(i) when person 1 jumps off :
m1u1 + m2u2 = m1V1 + m2V2

0 = 75 * 2 - 1225 * v2

v2 = 6/49 ms-1

(ii) when person 2 jumps off :
m1u1 + m2u2 = m1V1 + m2V2

0 = 75 * (2 + 6/49) - 1150 * v2

v2 = 156/1127 ms-1

(iii) when person 3 jumps off :
m1u1 + m2u2 = m1V1 + m2V2

0 = 75 * (2 + 156/1127) - 1075 * v2

v2 = 7230/48461 ms-1

(iv) when person 4 jumps off :
m1u1 + m2u2 = m1V1 + m2V2

0 = 75 * (2 + 7230/48461) - 1000*v2

v2 = 0.161 ms-1

I calculated this with respect to the car. Do I get it right? For this kind of question, what should be taken as the reference, the car or the ground?

Thanks

EDIT : I just realized my mistake. The reference should be ground and when person 2, 3 and 4 jump off one by one, there should be initial momentum relative to the ground. I'll amend my calculation in the next post

Last edited: Feb 7, 2010
2. Feb 7, 2010

### songoku

a. the answer is still the same

b. assume the car moves to the left and the person jumps off to the right
(i) when person 1 jumps off :
0 = 75 * 2 - 1225 * v2

v2 = 6/49 ms-1

(ii) when person 2 jumps off :
-(75 * 6/49) - 1225 * 6/49 = 75 * (2 + 6/49) - 1150 * v2

v2 = 312/1127 ms-1

(iii)when person 3 jumps off :
- 75*312/1127 - 1150 * 312/1127 = 75*(2 + 312/1127) - 1075 * v2

v2 = 0.474 ms-1

(iv)when person 4 jumps off :
- 75 * 0.474 - 1075 * 0.474 = 75 * (2 + 0.474) - 1000 * v2

v2 = 0.73 ms-1

So the final speed of the car = 0.73 ms-1

Do I get it right? Thanks

Last edited: Feb 7, 2010
3. Feb 8, 2010

### songoku

Sorry fo bumping...

Can someone please check my work? Thanks

4. Feb 8, 2010

### dr_k

I agree with this part. Here's my picture (jumping off to the left):
http://img695.imageshack.us/img695/9197/mv1.jpg [Broken]

I don't agree with these. My picture, for part (ii) is,
http://img246.imageshack.us/img246/6475/mv2q.jpg [Broken]

when I use a frame at rest, although, it would be easier to make a new inertial frame for each jump, that moves at the constant speed of the previous jump. Basically, you can use a frame at rest to get $\rm V_1$, then a frame moving at $\rm V_1$ to get $\rm V_2$,...etc.

The jumpers each give $\rm \frac{mv}{3m+M}$, $\rm \frac{mv}{2m+M}$, $\rm \frac{mv}{m+M}$, and $\rm \frac{mv}{M}$ respectiviely, which can be summed to find the final speed.

Last edited by a moderator: May 4, 2017
5. Feb 8, 2010

### songoku

Hi dr_k

So for the case when person 2 jumps off, the equation should be (based on your picture):

(3*75 + 1000) * 6/49 = - (2 - 6/49) * 75 + 1150 * V2 ??

And I don't understand about this part "it would be easier to make a new inertial frame for each jump, that moves at the constant speed of the previous jump. Basically, you can use a frame at rest to get V1 , then a frame moving at V1 to get V2, ... etc"

Can you please explain more? Thanks

6. Feb 8, 2010

### dr_k

Yep. Drawing the picture is 1/2 the problem.

Well, solving for $\rm V_2$, in the second picture, gives $\rm V_2 = V_1 + \frac{m v}{2 m + M}$, where $\rm V_1 = \frac{m v}{3 m + M}$. Basically the objects in the the (i) initial and (f) final picture are all traveling at $\rm V_1$. Instead of carrying these terms, you could just make an inertial frame moving at constant speed $\rm V_1$, and then remember to sum up the speed of the frames at the end. For example, the second picture above now looks like

http://img199.imageshack.us/img199/4534/frames1.jpg [Broken]

This frame says $\rm V' = \frac{m v}{2 m + M}$. To get the speed for an observer at rest, just add $\rm V_1 + V'$ . Now do this as many times as required to get all four masses off.

Last edited by a moderator: May 4, 2017
7. Feb 12, 2010

### songoku

Hi dr_k

Maybe I get your point. I'll try it first. Thanks a lot !!

8. Feb 13, 2010

### dr_k

Good Morning Songoku,

I was just looking at this problem, and I noticed that I didn't read the problem carefully enough. There is a subtle point that I neglected. The problem states "The people jump off the car with speed 2 ms-1 relative to the car". Relative to the car implies that I missed something. In part "(a) a. simultaneously", the picture should look like this:

http://img18.imageshack.us/img18/4534/frames1.jpg [Broken]

This picture correctly shows both 4m and M moving with $\rm V_1$, and therefore v is "relative" to the car.

This implies

$\rm 0 = M V_1 - 4m (v-V_1)$

so

$\rm V_1 = \frac{4mv}{M+4m}$ .

Putting the numbers in gives $\rm V_1 = .4615 m/s$ for part (a).

Give me a few minutes, and I'll write something up for part (b).

Last edited by a moderator: May 4, 2017
9. Feb 13, 2010

### dr_k

For part (b):

http://img688.imageshack.us/img688/7726/frames2.jpg [Broken]

Here v is "relative" to the car, and we get

$\rm V_1 = \frac{mv}{4m+M}$

Now I make a new inertial frame, moving in a straight line at constant speed $\rm V_1$ ,

http://img708.imageshack.us/img708/8674/frames3.jpg [Broken]

where

$\rm V_2 = \frac{mv}{3m+M}$

Do this two more times, with new inertial frames, to get the last two masses off. The final speed of the car is therefore

$\rm V_{car} = V_1+V_2+V_3+V_4 = \frac{mv}{4m+M}+\frac{mv}{3m+M}+\frac{mv}{2m+M}+\frac{mv}{m+M} = .5078 m/s$

Last edited by a moderator: May 4, 2017
10. Feb 22, 2010

### songoku

Hi dr k

Ah I see my mistake for neglecting "relative" . Thanks !!

11. Feb 22, 2010

### dr_k

I missed it as well! But I think we've got it covered now.