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.

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