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# Mechanics Help watch

1. A ball of mass 0.2kg is dropped from a height of 2.5m above horizontal ground. After hitting the ground it rises to a height of 1.8m above the ground, Find the magnitude of the impulse received by the ball from the ground.

2. alright, well impulse is change in momentum, so the ball's velocities immediately before and after impact need to be worked out. you can do this with suvat:
s=2.5, u=0, v=?, a=9.8, t=? for velocity beforehand

s=1.8, u=?, v=0, a=-9.8, t=? for velocity afterwards

hint: use v^2 = u^2 + 2as

Hopefully you can get it from there, but if not i'll help a little more, depends how lazy you are and how much you actually wanna learn how to do it or just get it done. quote me for help.
3. Find the velocity at which the ball hit the ground and the velocity with which the ball bounced up. Then it's just substituting the values into
4. Ok Tweek, you should have learnt that the Impulse a body experiences is equal to its change in momentum. so

where m is the mass of the body and u and v the inital and final veleocites respectivly. When is there a change of momentum in the question? Now what of these variables do you already now. Hint, what equation do you know to calculate the velocity of a body under constant acceleration. Also consider if a body was simply thrown from the ground and raised a height of 1.8m, what would its initial velocity have to have been?
5. Ok thanx
But i keep on getting impluse as 0.22Ns
but the answer supposedly is 2.59 Ns
6. Always use:
* The direction of Impulse exerter as the positive (for velocity)....
* And mass and velocity of the impulse reciever....

Try the question now.

Final Equation is
I = m (v - u)
I = 0.2 ( 5.9 - - 7) = 0.2 ( 5.9 + 7) = 2.59 Ns

Keep in mind that
v upon hitting ground = sqrt (u^2 + 2as)
v upon leaving ground = v upon falling from displacement of bounce = sqrt (u^2 + 2as)
7. (Original post by Tweek_1992)
Ok thanx
But i keep on getting impluse as 0.22Ns
but the answer supposedly is 2.59 Ns
Can you show us your workings?

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-2.5 = 0 - 0.5*9.8*t^2

t = 5/7

v = u + 9.8*5/7

v = 7

0 = u^2 - 2*9.8*1.8

u = -5.939 (the positive direction is downwards, the ball goes up so it has negative velocity)

I = 0.2 (-5.939 -7) = - 2.59

|I| = 2.59 Ns
8. i found the finial velocity by SUVAT as 5.9ms^-1 and the initial as 7ms^-1
then i used the formula
I=mv-mu
I=(0.2*7)-(0.2*5.9)
I=0.22
9. (Original post by Tweek_1992)
i found the finial velocity by SUVAT as 5.9ms^-1 and the initial as 7ms^-1
then i used the formula
I=mv-mu
I=(0.2*7)-(0.2*5.9)
I=0.22
Son,
Your mistake is in plugging in to equation.
Formula is I = m(v-u)
Taking positive to exerters exerting direction.
v = 5.9
u = -7
Hence:
I = m (v - u)
I = 0.2 ( 5.9 - - 7) = 0.2 ( 5.9 + 7) = 2.59 Ns

Which I now avoid by:

Using:
* Taking Impulse exerter's exerting direction as the positive direction (for velocities)....
* And mass and velocity of the impulse reciever....
10. (Original post by Tweek_1992)
i found the finial velocity by SUVAT as 5.9ms^-1 and the initial as 7ms^-1
then i used the formula
I=mv-mu
I=(0.2*7)-(0.2*5.9)
I=0.22
The final velocity is -5.9 (the ball is going upwards, whereas the positive direction is downwards.

And I = m(v-u), not m(u-v)

So I = 0.2 (-5.9 - 7) = -2.59

Magnitude is always positive though, so |I| = 2.59

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