Another Conservation of momentum Problem

A steel ball of mass 40.0g is dropped from a height 2.00m, it rebounds 1.60m what is the impluse during the time of contact.


mgh_1 + (1/2)m(v_i^2) = mgh_2 + (1/2)m(v_f^2)

mgh_1 = (1/2)m(v_f^2)
2gh_1 = v_f^2

v_f = sqrt(2gh_1), where h1 = 2.00m


Now impluse is the change in momentum so:

J = p_2 - p_1 ->  J = 0 - m*sqrt(2gh_1)

Right, since the ball is stopped at 1.6m the final momentum has to be zero? However, I dont think this is the right answer.


Brian
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BrianGEFF719Asked:
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ozoCommented:
the ball and the earth exchange momentum through gravitational interactions, and through impact.
just before impact, v_1=sqrt(2gh_1)
just after impact, v_2=-sqrt(2gh_2)
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borislavmarkovCommented:
E_initial = E_contact = m.g.h = mv^2/2
v=sqrt(2g.h)

p = mv = 0.040 * sqrt(2*9.8*2) = 0.250 [kg.m/s]

The rebound has nothing to do with solution, this is a trick.
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BrianGEFF719Author Commented:
borislavmarkov : we are trying to find the impluse during the time of contact....not the momentum.
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BrianGEFF719Author Commented:
>>the ball and the earth exchange momentum through gravitational interactions, and through impact.

I dont understand what you mean ozo, the way I see it is the ball has some velocity before it hits the steel plate. The velocity it has is sqrt[2gh].

Now the ball is going to rebound until it has no velocity right?

I'm very confused about this whole momentum thing.
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ozoCommented:
> the ball is stopped at 1.6m the final momentum has to be zero
but that doesn't mean the momentum at 0m has to be zero, because the earth and the ball can exchange momentum with each other while the ball is between 0m and 1.6m

The spring is going to compress until the ball has no velocity in the middle of the impact,
then the spring will expand until the contact ends with the ball having sufficient velocity to rise to 1.6m
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BrianGEFF719Author Commented:
>>because the earth and the ball can exchange momentum with each other while the ball is between 0m and 1.6m

What do you mean by this, how can the earth and the ball exchange momentum?
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ozoCommented:
Gravity can act between the ball and the earth.
The ball pulls on the earth just as hard as the earth pulls on the ball,
so like the rifle and the bullet, any momentum that one gets is balanced by
the opposite momentum change in the other.
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borislavmarkovCommented:
sorry
here is second try:
p1 = mv1 = 0.040 * sqrt(2*9.8*2) = 0,25044
p2 = -mv2 = 0.040 * sqrt(2*9.8*1.6) = 0,224

p2 has opposite direction:
Impulse = Change in Momentum
J = DP = -p2-p1 = -0,47444 [N.s]

I think this should be correct.

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BrianGEFF719Author Commented:
>>The spring is going to compress until the ball has no velocity in the middle of the impact,
>>then the spring will expand until the contact ends with the ball having sufficient velocity to rise to 1.6m

Sorry this is taking so much time ozo.

But The spring will expand until the contact ends with the ball having sufficent velocity to rise to 1.6m. So the velocity of the ball at 1.6m is zero.

Momentum = mass * velocity

P = m * 0 = 0

I still dont quite understand how an object reaching a maximum height as a projectile could have momentum when its velocity is zero because momentum depends on velocity.

Brian
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ozoCommented:
Its momentum at the point wnen it rebounds 1.60m is 0,
but I think the question is asking about what happened 1.60m earlier, at the end of its contact with the ground.

The momentum was also 0 when it started at 2.00m, but it was not 0  at the first moment of contact.
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borislavmarkovCommented:
Ozo, Brian, did you looked at my calculations. Don't they seem to be correct?
Velocity at the moment before rebound is pointing to the ground.
Velocity at the moment after rebound is pointing up, so the moment direction is changed. So it looks to be negative. Why you don't accept my answer?
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