Speed of car related to momentum

In summary, 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. The speed of the car is 0.6 ms-1 if the people jump off simultaneously. If the people jump off one by one, the final speed of the car is 0.73 ms-1.
  • #1
songoku
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Homework Statement


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

Homework Equations


conservation of momentum

The Attempt at a Solution



a.
m1u1 + m2u2 = m1V1 + m2V2

0 = 300 * 2 - 1000 * v2

v2 = 0.6 ms-1b.
(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?

ThanksEDIT : 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
 
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  • #2
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
 
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  • #3
Sorry fo bumping...

Can someone please check my work? Thanks :smile:
 
  • #4
songoku said:
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

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




(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

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 [itex] \rm V_1[/itex], then a frame moving at [itex] \rm V_1[/itex] to get [itex] \rm V_2[/itex],...etc.

The jumpers each give [itex]\rm \frac{mv}{3m+M}[/itex], [itex]\rm \frac{mv}{2m+M}[/itex], [itex]\rm \frac{mv}{m+M}[/itex], and [itex]\rm \frac{mv}{M}[/itex] respectiviely, which can be summed to find the final speed.
 
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  • #5
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
songoku said:
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 ??

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

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

Well, solving for [itex] \rm V_2[/itex], in the second picture, gives [itex] \rm V_2 = V_1 + \frac{m v}{2 m + M}[/itex], where [itex] \rm V_1 = \frac{m v}{3 m + M}[/itex]. Basically the objects in the the (i) initial and (f) final picture are all traveling at [itex] \rm V_1[/itex]. Instead of carrying these terms, you could just make an inertial frame moving at constant speed [itex] \rm V_1 [/itex], 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 [itex] \rm V' = \frac{m v}{2 m + M} [/itex]. To get the speed for an observer at rest, just add [itex] \rm V_1 + V' [/itex] . Now do this as many times as required to get all four masses off.
 
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  • #7
Hi dr_k

Maybe I get your point. I'll try it first. Thanks a lot ! :smile:
 
  • #8
songoku said:

Homework Statement


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


Homework Equations


conservation of momentum


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

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 [itex] \rm V_1[/itex], and therefore v is "relative" to the car.

This implies

[itex] \rm 0 = M V_1 - 4m (v-V_1)[/itex]

so

[itex] \rm V_1 = \frac{4mv}{M+4m}[/itex] .

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

Give me a few minutes, and I'll write something up for part (b).
 
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  • #9
For part (b):

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

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

[itex] \rm V_1 = \frac{mv}{4m+M}[/itex]

Now I make a new inertial frame, moving in a straight line at constant speed [itex] \rm V_1[/itex] ,

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

where

[itex] \rm V_2 = \frac{mv}{3m+M}[/itex]

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

[itex] \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[/itex]
 
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  • #10
Hi dr k

Ah I see my mistake for neglecting "relative" . Thanks ! :smile:
 
  • #11
songoku said:
Hi dr k

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

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

1. What is the relationship between the speed of a car and its momentum?

The speed of a car and its momentum are directly proportional. This means that as the speed of the car increases, its momentum also increases. Similarly, if the speed decreases, the momentum decreases.

2. How does the mass of a car affect its momentum?

The mass of a car has a significant impact on its momentum. A car with a larger mass will have a greater momentum compared to a car with a smaller mass, even if they are travelling at the same speed.

3. Can a car have a high speed but low momentum?

Yes, it is possible for a car to have a high speed but low momentum. This can happen if the car has a small mass. In this case, the high speed is not enough to compensate for the low mass, resulting in a lower momentum.

4. How does the direction of motion affect the momentum of a car?

The direction of motion does not have a direct effect on the momentum of a car. However, it does affect the direction of the momentum vector. For example, if a car is moving east with a certain momentum, changing its direction to north will result in a change in the direction of its momentum vector.

5. How is the speed of a car related to its kinetic energy?

The speed of a car is directly related to its kinetic energy. This means that as the speed increases, the kinetic energy also increases. The relationship between speed and kinetic energy is defined by the equation KE = 1/2 * m * v^2, where m is the mass of the car and v is its speed.

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