Question

Impulse Question!!!!

Answer 1

Q2 A 0.80-kg ball is dropped from a height of 2.0 m above the
ground. It rebounds to a height of 1.6 m. The ball is in contact
with the ground for 0.050 s. (i) Determine the impulse exerted
by the ground (and gravity) on the ball in the 0.050 s.
(ii) Determine the average net force on the ball in the 0.050 s.

1.How would I solve for the impulse exerted by the ground?
2. For my net force my answer was -6.0 am I correct?

Answer 2

My net force should've been -6.4 and not 6.0 if I have done this the correct way.

Answer 3

\[\huge J=F\Delta t \implies J=mg*0.05\text{s}\]I think this equation will work for a).

Answer 4

I feel like the net force would be 0 while the ball is in contact with the surface. If we were to draw a free body diagram, we would have the normal force and the weight of the ball. But since those are our only components, and they are in opposite directions, then that means that the normal force = weight, or in other words, the normal force - weight = 0.

Answer 5

Is -6.4N the correct answer or is it your answer?

Answer 6

Use conservation of energy to find the velocity of the ball right before it hits the ground and right after it hits the ground. Use these velocities to find the change in momentum. Then use the known time interval to find the average force exerted to cause the change in momentum

Answer 7

because the collision is not totally elastic, there is energy loss and you can't use conservation of energy between the two end states. However, you CAN use conservation of energy from 2m to 0m and from 0m to 1.5m. The loss in between the two states could be due to things like sound, deformation, heat etc. I'd be happy to explain more if needed!

Answer 8

Impulse is a change in momentum and is expressed in N.s.

\(\vec I=\Delta \vec p\)