Impulse & Change in Momentum: 0.5kg Object Dropping 3m

[crafty2288]

Homework Statement

A 0.5 kg object drops from rest from a 3m high point. On bouncing from Earth it rises to a height of 2m. The collision with Earth lasts for about 10^-4 seconds.

i) What is the impulse received by the object?
ii)Considering that there is a gravitational force present, Can you assume that in the collision F(ext)=0? If so, what is the change in the Earth's momentum?
iii)Is this collision totally elastic? Justify.

Homework Equations

J=F*(delta)t
F=ma

The Attempt at a Solution

i)So, I've got the first part down. Impulse will be the mass*acceleration*change in time. Which is (.5)(9.8)(10^-4) = 0.00049

This is where I need help:
ii)I'm not sure if this can be assumed. If there IS a change in the Earth's momentum and F(ext)=0, it would just be -0.00049... But I'm unsure about this part. Help?

iii) If this was totally elastic, the ball would return to 3m high, not 2m - right? So this is NOT totally elastic. (Let me know if I'm confused here.)

Thank you in advance for helping me to understand this.

 

[pat666]

OK so first of the velocity at the point on impact:
mgh=1/2mv^2 9.81*3*2=v^2
V=7.67m/s
next v instantly after collision:
0=u^2-2*9.81*2
u=6.26

from that change in momentum
7.67*.5-6.25*.5=0.704

so J=0.704that is what i get for impulse, are you sure your answer is right?

 

[pat666]

to solve this problem we are assuming that there is no external forces. however the Earth's momentum is not changed (from this collision) as the momentum is conserved in the system.
Part 3 is correct i think since KE would be conserved.
Can you please let me know the answer your book gives for impulse because I don't think that your answer is correct.

 

[crafty2288]

OK so first of the velocity at the point on impact:
mgh=1/2mv^2 9.81*3*2=v^2
V=7.67m/s
next v instantly after collision:
0=u^2-2*9.81*2
u=6.26

from that change in momentum
7.67*.5-6.25*.5=0.704

so J=0.704that is what i get for impulse, are you sure your answer is right?

No, I am not sure... but I don't understand half of what you're doing there.

What is this? mgh=1/2mv^2 1/2mv^2 is Kinetic Energy, but why does that equal mgh? How are you finding the velocity after the collision? Can you break that down a little...? Show more steps? I'm really not following you at all.

Can you please let me know the answer your book gives for impulse because I don't think that your answer is correct.

I don't have an answer sheet. It's a study guide for an exam. The professor just gives us a packet of 50+ problems that can potentially show up on the exam, and leaves us to fend for ourselves. I have no way of checking.

 

[pat666]

ok. so mgh is the potential energy of any object at a height- this is equal to the kinetic energy at the bottom, you could use kinematic equations for this but i find energy conservation easier. therefore mgh=1/2mv^2 = 2gh=v^2

for the velocity after the collision.
it must reach a height of 2m (given)
so v^2=u^2+2as (kinematic equation where v=0, u is what i want, a is -9.81 and s=2)

impulse is equal to the change in momentum
so J=mV1-mV2ok so where i think you've gone wrong is you've used a static acceleration for your impulse. however, over that 5seconds the acceleration not only changes value but completely changes direction. the value for time is not used and i think they put it there to confuse people (apparently it worked)

hope this was helpful?let me know if you understand it now and if it was helpful

 

[crafty2288]

My understanding of the definition of Impulse is that its the average amount of force applied over an amount of time, though. How can you find Impulse without using a change in time?

 

[pat666]

it does use time without using time--- you should have been taught that impulse is equal to the difference in momentum (that is known fact i didn't just make it up) which is relative to the contact time but you only need the "extreme" values. your answer was way to small impulse is how many Newtons of force is transferred.