## amishra Group Title 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. one year ago one year ago

1. CliffSedge Group Title

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

2. amishra Group Title

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

3. CliffSedge Group Title

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

4. CliffSedge Group Title

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

5. Yahoo! Group Title

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

6. amishra Group Title

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. Yahoo! Group Title

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

8. Zekarias Group Title

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

9. CliffSedge Group Title

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

10. CliffSedge Group Title

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

11. Yahoo! Group Title

That is Impulse @CliffSedge

12. CliffSedge Group Title

Impulse divided by time = force.

13. amishra Group Title

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

14. CliffSedge Group Title

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

15. CliffSedge Group Title

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

16. Yahoo! Group Title

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

17. Yahoo! Group Title

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

18. CliffSedge Group Title

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

19. Yahoo! Group Title

20. amishra Group Title

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

21. CliffSedge Group Title

I already provided it (in formula form, anyway)

22. Yahoo! Group Title

23. amishra Group Title

6 Newtons

24. CliffSedge Group Title

That's what I got too. :-)

25. CliffSedge Group Title

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

26. CliffSedge Group Title

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. amishra Group Title

@CliffSedge Oh, thank you for sharing that :D