How is Impulse Related to Momentum?

  • Thread starter Thread starter pirateboy
  • Start date Start date
  • Tags Tags
    Impulse Momentum
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
pirateboy
Messages
2
Reaction score
0
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 [tex]\vec{v}_0 = (20 \text{m}/\text{s})\hat{i}[/tex] is accelerated by a varying net force [tex]\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}[/tex] over a time interval of 3.0 seconds.

a) What is the net implulse [tex](\vec{J})[/tex] 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


[tex]\vec{p} = m\vec{v}[/tex]

[tex]\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}[/tex]

[tex]\vec{J} = \int_{t_i}^{t_f}\vec{F}(t)\,dt[/tex]

[tex]F_{\text{avg}} = \frac{J}{\Delta t}[/tex]

The Attempt at a Solution

I had trouble at part b.

a)So [tex]\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}[/tex]

b)This is where I was confused.

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

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

c)
[tex]F_{\text{avg}} = \frac{39}{3.0-0} = 13\,\text{N}[/tex]
 
Last edited:
Physics news on Phys.org
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.
 
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!