# Quick Mechanics 1 question

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1. Aug 17, 2015

### Charles W

1. The problem statement, all variables and given/known data

Ike and Jim are sitting in toy trucks; the masses are 60kg and 40kg respectively. The trucks are moving at 8 metres per second along a track, with Ike's behind Jim's. Ike pushes Jim's truck away with a pole, and Jim moves off 2 metres per second faster than Ike. What is Jim's new speed?

2. Relevant equations

Ft = mv - mu

3. The attempt at a solution

I have tried using Conservation of Momentum and Newton Third's Law, but cannot seem to come to a correct solution.

Thanks!

2. Aug 17, 2015

### SammyS

Staff Emeritus
What did you get using Conservation of Momentum and Newton Third's Law ?

3. Aug 17, 2015

### SteamKing

Staff Emeritus
Hi Charles:

Welcome to PF! We ask that members seeking help with HW problems post their questions in the appropriate HW form. This helps get relevant answers to your request fro help. This is why your post has been moved from the Pre-calculus HW forum.

Good Luck.

4. Aug 17, 2015

### Orodruin

Staff Emeritus
You need to actually show us what you did, not just state that you have done it, how else are we supposed to see where you go wrong?

5. Aug 17, 2015

### Charles W

I wrote down two expressions:

60u + 40v = 800kg metres per second (using the equation for momentum to work out 800 kg metres per second - (60*8) + (40*8)
v = u + 2 metres per second

Sorry for not providing enough information, but after this stage I could not think of any other approaches

6. Aug 17, 2015

### BvU

So instead of v you write u+2 in that equation. That leaves one equation with u as the only unknown ....

7. Aug 17, 2015

### Charles W

So: 60u + 40(u+2) = 800kg metres per second
Then I can simplify this to 100u + 40 = 800 kg meters per second

Then I'm not sure where to go from here because the units don't seem to work for a subtraction?

8. Aug 17, 2015

### Orodruin

Staff Emeritus
That would be because you have been far too liberal with your use of units on the left-hand side ...

9. Aug 18, 2015

### Charles W

Is 60u and 40u in kg metres per second as well?

10. Aug 18, 2015

### BvU

Simply write out completely the variables involved (so: value and dimension) :
becomes
60 kg * u m/s + 40 kg * (u+2) m/s = (60+ 40) kg * 8 m/s

Do this until you're fluent and confident with it (and longer if you're smart).

11. Aug 18, 2015

### Charles W

Thank you - I can see the logic of splitting it up, but I cannot see how I am able to deduce the value of u from this?

12. Aug 18, 2015

### BvU

You manipulate this until you have all knowns on the righthand side and u m/s on the left hand:
60 kg * u m/s + 40 kg * (u+2) m/s = (60+ 40) kg * 8 m/s
60 kg * u m/s + 40 kg * u m/s + 40 kg * 2 m/s = 100 kg * 8 m/s
(60 kg + 40 kg) * u m/s + 40 kg * 2 m/s = 100 * 8 kg m/s
and so on. It's like solving $60 x + 40 (x+2) = 800$ but a bit more extensive....

13. Aug 18, 2015

### Charles W

Oh, I understand now! Thank you very much for your help :)