# Momentum Problem on a particle

1. Feb 29, 2008

### Nightrider55

1. The problem statement, all variables and given/known data
Force F_x =(10N){sin({(2pi(t))/4.0s) (where t in (m/s) is exerted on a 420 g particle during the interval 0 less than or equal to T is less than or equal to 2 seconds.

If the particle starts at rest what is its speed at t=2 seconds?
2. Relevant equations

Jx= area under the Fx(t) cureve between Ti and Tf

Pfi-Pix=Jx

3. The attempt at a solution
This problem looks fairly easy but I am having trouble setting it up. I took the integral of the force over the time interval and got .54817 Ns which should be the impulse. I then used this to find Pfx by the equation above. I then used Vfx=(Pfx/m) to find the final velocity which came out to be 2.19 m/s but its wrong, any insight as to what I did wrong?

2. Feb 29, 2008

### jun9008

integral of F(t)= impulse= change of momentum= Final momentum, since intial velocity is zero.
Final momentum devide by the mass, .420kg, would equal the velocity.

I do not have a calculator right now but I am sure this is it :)

oh by the way, when integrating force, you might want to include absolute value around F(t)

Last edited: Feb 29, 2008
3. Feb 29, 2008

### Nightrider55

It keeps on saying my answer is wrong. What am I missing?

4. Feb 29, 2008

Use this:
$$F=m\frac{dv}{dt}$$

You know integration, right ?

5. Feb 29, 2008

### Nightrider55

Yea I think Im doing it right, after I integrate it I get 12.73 then you say times that by the mass, so 12.73 x .42kg= 5.35. That should be right, could someone double check.

6. Feb 29, 2008

### Dick

I integrated Fdt from t=0 to t=2 and got (40/pi)*N*sec. Why are you getting .54817Ns? Can you show details?

7. Feb 29, 2008

I got this on integrating:
$$0.042v=-\frac{2}{\pi}cos(\frac{\pi t}{2})+\frac{2}{\pi}$$

for t=2 it gives v=30.33 m/s.

8. Feb 29, 2008

### Dick

Ok, impulse=40/pi*Ns=12.73*Ns=m*v. That's good. Don't you want to divide the impulse by the mass to get velocity?

Last edited: Feb 29, 2008