# Question on Impulse-Momentum theory

A 5kg stone kicked by a student undergoes a velocity change from 6m/s to 10m/s in 4 seconds;

---- What is the momentum change of the stone within the time interval?

---- What impulse was impacted on the body during this time interval?

For the first question : $$p_2-p_1= M V_1 - MV_2 = M(V_1-V_2)= 5\textrm{ kg}(10-6)\textrm{m s}^{-1} = 20 \textrm{ kg m s}^{-1}$$

Is that correct for the first question?

As for the second : I'm stuck. Help.

$$\textrm{impulse}=I = F\cdot t$$ $$F = (mv-mu)/t$$ $$F = m(v-u)/t$$ $$F\cdot t = m(v-u)$$ $$\Rightarrow I = 5\textrm{ kg}(10-6)\textrm{ m s}^{-1} = 20 \textrm{ kg m s}^{-1}$$

is that correct? Because there has been no need to use time (4 seconds)......

or : $$F = ma = 5 \textrm{ kg} \frac{(10-6)\textrm{ m s}^{-1}}{4\textrm{s}}= 5\textrm{ N}$$

$$\textrm{impulse}= I = F\cdot T= 5\textrm{ N}\cdot 4 \textrm{ s} = 20\textrm{ N s},$$

still the same 20.

Thanks

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Refined. Retract downvote. –  Mob May 1 '12 at 15:15
I would suggest to use LaTeX markup for styling your derivations. If you want, I can edit your question for a better readability. –  Brent May 1 '12 at 15:57
@Brent: go ahead, that would be useful. (For just cosmetic edits like that, I usually don't bother to ask, since it's definitely an improvement.) –  David Z May 1 '12 at 19:14
It seems the homework tag applies even if it is not actual homework, cf. the tag description physics.stackexchange.com/tags/homework/info –  Qmechanic May 3 '12 at 8:52
Yes, your thinking is correct. The impulse imparted is equal to both the applied force times the time and to the change in momentum, $$I=F\Delta t=\Delta p.$$ (This is exactly Newton's second law.) The time taken by the kick is superfluous in your question.