## anonymous 3 years ago A ball of mass 0.5kg is thrown vertically down at 12m/s onto a driveway. If it is in contact with the driveway for 1.5 seconds, and rebounds with a velocity of 6m/s, find the average force of the driveway on the ball.

1. anonymous

You can use impulse-change-in-momentum relation.

2. anonymous

I used it but I didn't get the answer. Can you show me?

3. anonymous

*sorry, 'impulse = change-in-momentum relation'

4. anonymous

Sure, what did you get for the change in momentum?

5. anonymous

F = change in momentum 0.5*12 - (-6*0.5)

6. anonymous

I took the change in velocity to be 6 m/s. And then I multiplied mass into this change in velocity. So I got 3 as my momentum. :/

7. anonymous

Note:...the Direction is opposite..)

8. anonymous

Use this formula $F_{av}{\Delta}t=m{\Delta}v$

9. anonymous

Yes, the change in velocity is 12+6 since it changes direction.

10. anonymous

That change in momentum divided by the time of contact (1.5s,really? That's a long contact time . . .) = the average force.

11. anonymous

That is Impulse @CliffSedge

12. anonymous

Impulse divided by time = force.

13. anonymous

Oh THANK YOU ALL! I got it now! I didn't see the change in direction! :D

14. anonymous

$\large impulse=F_{avg} \cdot \Delta t$

15. anonymous

$\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}$

16. anonymous

lol...it is same as Finding change in momentum

17. anonymous

$F=\frac{ dP }{ dT }$

18. anonymous

Correct, but the question asked for average, not instantaneous.

19. anonymous

20. anonymous

I'm sorry, I used the impulse formula, and I got the correct answer. :) It doesn't really make a difference.

21. anonymous

I already provided it (in formula form, anyway)

22. anonymous

23. anonymous

6 Newtons

24. anonymous

That's what I got too. :-)

25. anonymous

$\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}$

26. anonymous

I also recommend getting in the habit of checking the units in your formulas. $\large N=kg\frac{m}{s^2}$ $\large kg\frac{m}{s^2} = \frac{(kg)(m/s)}{(s)}$

27. anonymous

@CliffSedge Oh, thank you for sharing that :D