# Work done by impulsive force

1. Oct 24, 2014

### erisedk

1. The problem statement, all variables and given/known data
A force exerts an impulse J on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is--
Ans: Ju/2

2. Relevant equations
Impulse= force*time
impulse=change in momentum

3. The attempt at a solution
Since applied force and initial velocity are oppositely directed, J=m(2u-(-u))=3mu.
Work=Force*displacement=Impulse*displacement/time=Impulse*velocity
But velocity is not constant. I don't know what to do from here.

2. Oct 24, 2014

### vela

Staff Emeritus
Hint: You can express the kinetic energy in terms of momentum: $K = \frac{p^2}{2m}$.

3. Oct 24, 2014

### haruspex

There's a bit of a catch here. Consider two phases. In the first phase, velocity changes from u in one direction to u in the opposite direction. How much work has been done so far?

4. Oct 24, 2014

### erisedk

Zero!oo) because change in kinetic energy is zero. Then change in ke when the velocity becomes 2u will be 1/2m(4u^2)-1/2mu^2=3/2mu^2. Impulse is 3mu so work is Ju/2!! Yay! Thanks for helping haruspex and vela. But now that I think about it, there is no need to consider two phases of the motion. All I can do is subtract initial ke from final ke and express it in the form of impulse and u.

5. Oct 25, 2014

### haruspex

Yes, but I thought you might fall into the trap of applying vela's equation to the change in momentum, giving K = $\frac{J^2}{2m} = \frac J{6u}$. The point being that the difference of the squares is not the same as the square of the difference.