Solving Momentum and Newton's Law: Ike and Jim's Truck Physics Problem

[Charles W]

Homework Statement

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?

Homework Equations

Ft = mv - mu

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!

 

[SammyS]

Homework Statement

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?

Homework Equations

Ft = mv - mu

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!

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

 

[SteamKing]

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.

 

[Orodruin]

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?

 

[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

 

[BvU]

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

 

[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?

 

[Orodruin]

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

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

 

[Charles W]

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

 

[BvU]

Simply write out completely the variables involved (so: value and dimension) :

60u + 40(u+2) = 800kg metres per second

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).

 

[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?

 

[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...

 

[Charles W]

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