# I Jumping off a moving cart on a frictionless track

1. Feb 28, 2017

### Nikhil Rajagopalan

If we consider a cart of mass M ,moving along a friction-less track, with velocity v on which a boy of mass m, stands at rest initially at the front end of the cart . If the boy jumps from the cart, in the direction of motion of the cart with a velocity u relative to the cart, what will be the velocity of the boy as he emerges out of the cart.
1. Will it be simply v+u.OR
2. Should this be computed using momentum conservation as follows.

Total momentum before jump = (M+m) * v

Total momentum after the jump = M * vf + m * (u + vf)

Where vf is the final velocity of the cart after the jump has been executed.

Which is the right method.

2. Feb 28, 2017

### jbriggs444

[emphasis mine]

It depends.

How do you read the problem? Does the boy jump from the cart so that his velocity is u relative to the cart's initial velocity or so that his velocity is u relative to the cart's final velocity? Or should we read it so that his velocity is u relative to the cart's average velocity?

3. Feb 28, 2017

### Nikhil Rajagopalan

jbriggs444, thank you for responding. the problem just says that the boy jumps with a velocity u relative to the cart. I understand the ambiguity though. How can we approach the problem in two different ways.
1. Boy jumps with velocity u relative to carts initial velocity.
2. Boy jumps in such a way that his velocity after the jump is u relative to the final velocity of the cart.

4. Feb 28, 2017

### Nikhil Rajagopalan

I have seen problems where a gun fires a bullet with some velocity from a moving train or a vehicle. in such cases, final velocity of the bullet is usually the firing velocity plus the velocity of the vehicle. The cases where something is projected with a velocity from a moving body. My aim is to draw a generalization for such situations.

5. Feb 28, 2017

### jbriggs444

You've already provided solutions to both versions of the problem.

6. Feb 28, 2017

### Nikhil Rajagopalan

I believe that, for the case where the velocity is relative to the initial velocity of the cart,

Total momentum before jump = (M+m) * v

Total momentum after the jump = M * vf + m * (u + vf)

Where vf is the final velocity of the cart after the jump has been executed.

Equating both of these, vf and hence u + vf could be computed.

In the case where the velocity u is relative to the final velocity of the cart, how could the analysis be.

7. Feb 28, 2017

### jbriggs444

Keep your eyes on the prize. What are you trying to calculate?

8. Feb 28, 2017

### Nikhil Rajagopalan

jbriggs444,sorry, i think i got it the other way round.

In case of the velocity u being relative to the initial velocity v of the cart. Momentum conservation could be.

(M+m) * v = m * (v +u) + M * vf

In case of the velocity of the boy u being relative to the final velocity of the cart,

(M+m) * v = m * (u + vf) + M * vf

Am i still missing something

9. Feb 28, 2017

### Nikhil Rajagopalan

If my previous assumption is true,

In case of the velocity u being relative to the initial velocity v of the cart.The velocity of the boy will be v + u

In case of the velocity of the boy u being relative to the final velocity of the cart. The velocity of the by will be vf + u

10. Feb 28, 2017

### jbriggs444

Right. Just as you have it in your original post.
I think you mean u + vf. That is correct. But you do not know vf. So you would have to make a conservation of momentum calculation. As you were setting up to do in your original post.

11. Feb 28, 2017

### Nikhil Rajagopalan

Thank you so much jbriggs444,

Apologies for the mistake,i had edited it soon after i made the post.

1. So,In problems which involves firing a bullet from a moving train or a car, with the velocity of the bullet given as u, without any specific detail which of the above cases should we resort to. Or the case when a projectile is projected from a moving cart.

2. In the case where a cricketer or a bowler runs in and throws a ball, considering the run up speed as 'vr' and the speed generated by swinging the arm as 'va' , will the speed of the delivery be va + vr ?

Does the fact that in the first case, its an explosion that causes the momentum to be transferred and in the second case, its a contact force make any difference in choosing the method.

12. Feb 28, 2017

### Nikhil Rajagopalan

For example, if a gun is moving with a velocity v and it fires a bullet with velocity u, or to be more specific, the bullet leaves the barrel with a relative velocity of u. Shouldn't this u be the velocity relative to the final velocity of the gun. In that case, i believe that the effective velocity of the bullet will be less than v + u. It should rather be vf + u , where vf is the final velocity of the gun corrected after recoil. Is this consideration correct.

13. Feb 28, 2017

### jbriggs444

If I were quoting a muzzle velocity, it would be relative to the ground and would assume a gun initially at rest on a shooter's shoulder.

If you encounter a homework problem that specifies a velocity u but does not indicate a frame of reference for that velocity then you may have to read carefully, ask for clarification or make and state a clarifying assumption. There is no magic formula.

14. Feb 28, 2017

### Nikhil Rajagopalan

jbriggs444, thank you for your continued support.

Considering a gun with an original muzzle velocity (that corresponds to your description of muzzle velocity. considering gun to be stationary) v, suspended free to move, hypothetically firing without any external support, will the bullet leave with the same velocity v ?

15. Feb 28, 2017

### jbriggs444

You are comparing a gun hanging from strings, for instance, to a gun solidly clamped in a machine rest and asking whether the resulting bullet velocities will be the same?

No, they will be different. [Probably not by enough to matter to the target sitting in the cross hairs]

16. Feb 28, 2017

### Nikhil Rajagopalan

yes, jbriggs444.

Comparing jumping from a stationary cart on a friction-less track and a suspended gun firing, i believe both scenarios are similar.

One thing that makes me ponder still, is that fact that inside a gun, there is a momentary explosion, and in case of the cart there is a more prolonged contact force. In the case of the explosion inside a gun, from the site of explosion, the bullet thrusts forward and the gun moves forward more like an after effect, and therefore, motion of the gun, basically doesn't affect the launch of the bullet since the explosion is already over. But, in case of the cart, the execution of the jump is a slower process, if i intend to jump at a velocity u, from the start of the attempt to the completion of the attempt, the cart goes back and eventually reducing the speed with which i emerge from the cart to a value lesser than u. Does this make the two scenarios different.?

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