# Impulsive force question

1. Dec 28, 2011

### Samuelb88

1. The problem statement, all variables and given/known data
Suppose that on the time interval [0,h] we have a force acting along the x-axis with magnitude $0 \leq f(t)$ at time t, and f(0) = f(h) = 0. For a particle of mass m starting at rest, so that its distance x(t) from 0 satisfies 0 = x(0) = x'(0), show that $x'(h) = \int_0^h f(t) dt$.

3. The attempt at a solution
This is one of those problems where I swear I am right and the book is wrong.

Let P denote the particle. Then on the interval [0,h], the force f(t) acts on P along the x-axis so henceforth I will write f(t) = (f(t),0) = f(t). The second law tells me that mx'' = f(t). Hence
$$\int_0^h f(t) dt = \int_0^h m x'' dt = \int_0^h mv' dt = mv(h) - mv(0) = mv(h)$$

since v(0) = x'(0) = 0. So this seems to contradict what I was suppose to show except in the special case that m=1. I haven't really used my assumption that f(0) = 0 = f(h), but I can't think of how that would tie into this problem other than tell me that there is no net force acting on P at both t=0 and t=h. Would someone mind giving me some pointers on this one?

- Sam

2. Dec 28, 2011

### Redbelly98

Staff Emeritus
You are correct. It's likely this is a simple typo in the book, omitting the m from the solution, especially since they went to the trouble of saying "a particle of mass m".