# Change of momentum of a crate within a system

1. Mar 26, 2012

### junkie_ball

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

Ok i am working through some course work questions and have been asked to calculate the change of momentum of the 200kg crate. I have the following information after working through the previous questions

Mass of crate = 200Kg
Distance moved (horizontally) = 20m
time = 4.07s
Initial Velocity = 0 m/s
Final velocity = 9.82 m/s
Aceleration = 2.41 m/s/s

I realise i don't need all this information but wanted to include all the information i have gained about the system. I have also attached my initial free body diagram.

2. Relevant equations

Impulse = ft = mΔv

3. The attempt at a solution

So this is where i'm having difficulties i think i have got confused with all the information i have already gained about the system but my attempt is as below would appreciate any pointers in the right direction.

Ft = mΔv

F = mΔv/t

F = (200Kg x 9.82m/s)/4.07s

F = 482.56N
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Free Body Diagram.doc
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2. Mar 26, 2012

### Staff: Mentor

What's the initial momentum? What's the final momentum? What's the difference (change)?

3. Mar 28, 2012

### junkie_ball

Momentum = Mass x Velocity

Initial Momentum = 200 x 0 = 0
Final Momentum = 200 x 9.82 = 1964

So to summarise the initial momentum would be 0 kgm/s and the final momentum is 1964 kgm/s as such the change of moment would be 1964 kgm/s? so the change in momentum would be + 1964 kgm/s?

I thought i needed to use the impulse = force applied x time equation? is this not the change of momentum. Thanks for you help.

4. Mar 28, 2012

### Staff: Mentor

You could use the f*t version if you wanted to. You've provided the acceleration and time, and the force can be obtained from f=ma. But it amounts to the same thing, and really, by definition the impulse is the change in momentum so why not go directly there?

Last edited: Mar 28, 2012
5. Mar 28, 2012

### junkie_ball

Thanks for the reply. I see your point (why make life difficult! :D) i have just tried the force * time to work out the impulse and that confirms my answer for me. Thanks for the help