How to Solve the Impulse Problem with a 70-g Steel Ball in Contact with a Plate

[preluderacer]

Homework Statement

A 70-g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 2.00 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:

The Attempt at a Solution

Im not sure about a lot in this problem. Do I calculate the velocity has it falling for 1 sec or .75 sec or I shouldn't need to do that? I am so confused on this one.

 

[housemartin]

I don't get why do you have those 2.00 s here. What is balls velocity when it hits the plate? When it re-bounces? How is change in momentum related to to impulse?

 

[preluderacer]

So does it fall down in 1 sec? Is it then moving at 9.8 m/s. Does it then go up at 9.8 m/s?

 

[Doc Al]

Do I calculate the velocity has it falling for 1 sec or .75 sec or I shouldn't need to do that?

You're told that the total round trip time is 2.00 s and that the time in contact with the floor is 0.5 ms, so as far as computing the speed as it hits the floor you can safely neglect that 0.5 ms.

So does it fall down in 1 sec? Is it then moving at 9.8 m/s. Does it then go up at 9.8 m/s?

Yes.

 

[preluderacer]

so what I did was have (0.070kg(9.8m/s)-0.070kg(9.8m/s) / 0.005 s ? This seems very wrong to me

 

[Doc Al]

so what I did was have (0.070kg(9.8m/s)-0.070kg(9.8m/s) / 0.005 s ? This seems very wrong to me

Careful with signs. Momentum is a vector--direction matters.