How is Impulse Related to Momentum?

[pirateboy]

I'm a bit confused by this question. If you could help me check my work, that'd be great.
Thanks.

Homework Statement

A 40kg object with an initial velocity $$\vec{v}_0 = (20 \text{m}/\text{s})\hat{i}$$ is accelerated by a varying net force $$\vec{F}(t)=\left[ \left( 3.0 \text{N}/\text{s}^2\right)t^2 - \left( 4.0 \text{N}/\text{s}^2 \right)t + 10.0\text{N} \right]\hat{i}$$ over a time interval of 3.0 seconds.

a) What is the net implulse $$(\vec{J})$$ received by the object using the force equation above.

b) What is the final momentum of the object at the end of the time interval?

c) What is the average net force exerted on the object over the time interval?

Homework Equations

$$\vec{p} = m\vec{v}$$

$$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$$

$$\vec{J} = \int_{t_i}^{t_f}\vec{F}(t)\,dt$$

$$F_{\text{avg}} = \frac{J}{\Delta t}$$

The Attempt at a Solution

I had trouble at part b.

a)So $$\int_{0}^{3.0}\vec{F}(t)\,dt = \left[ \left(t^3 -2.0t^2+10.0t \right)\hat{i} \right]_{0}^{3.0} = 39 \text{kg}\cdot\text{m}/\text{s}$$

b)This is where I was confused.

Well, since $$\vec{p} = m\vec{v}$$, I solved the indefinate integral
$$\int\vec{F}(t)\,dt = t^3-2t^2+10t + C$$
And since
$$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$$ and $$\vec{p} = m\vec{v}$$
then
$$\vec{p}(0) = t^3 - 2.0t^2 + 10.0t + C = m\vec{v}_0 = 800$$,
so $$C = 800$$?

So final momentum is
$$\vec{p}(3) = (3)^3 -2(3)^2 + 10(3) + 800 = 815 \text{kg}\cdot\text{m}/\text{s}??$$

c)
$$F_{\text{avg}} = \frac{39}{3.0-0} = 13\,\text{N}$$

 

[Delphi51]

Welcome to PF, pirateboy.
I don't see why (b) needs to be so complicated.
The impulse is equal to the change in momentum, so your 39 is the change in momentum. Final momentum is 40*20 + 39.

I think you have an adding error in the line where you get 815. I get 839 there.

 

[pirateboy]

where did that 15 come from? that's what i get for having a jumbled scratch sheet, i guess.

and wow, i guess 40*20 + 39 does make sense, doesn't it. i guess i just got caught up in making things difficult for myself. thanks!

 

[Delphi51]

Most welcome!

 

[rwisz]

Your solution for part a is correct.

For part b, you are on the right track. The final momentum should be equal to the initial momentum plus the impulse received by the object. So, in this case, it would be:

\vec{p}(3) = (800 \text{kg}\cdot\text{m}/\text{s}) + (39 \text{kg}\cdot\text{m}/\text{s}) = 839 \text{kg}\cdot\text{m}/\text{s}

For part c, you have correctly calculated the average net force. Good job!