Balls and clay blob - Collision problems

[nns91]

Homework Statement

A ball of mass 9m is dropped from rest from a height H =5 meter above the ground. It undergoes a perfectly elastic collision with the ground and rebounds. At the instant that the ball rebounds, a small blob of clay of mass m is released from rest from the original height H, directly above the ball. The clay blob, which is descending, eventually collides with the ball, which is ascending. Assume that g=10 m/s^2, that air resistance is negligible, and the collision process takes negligible time

a. Determine the speed of the ball immediately before it hits the ground.
b. Determine the time after the release of the clay blob at which the collision takes place. c. Determine the height above the ground at which the collision takes place.
d. Determine the speeds of the ball and the clay blob immediately before the collision.
e. If the ball and the clay blob stick together on impact, what is the magnitude and direction of their velocity immediately after the collision

Homework Equations

Ei=Ef
Pi=Pf

The Attempt at a Solution

I solved a using conservation of energy.

Other parts I cannot solve. Can anyone help ??

 

[nns91]

In part b, do I have to calculate the height at which the collision happens or is there any other way I can calculate the time ?

 

[Delphi51]

That sounds good, nns - perhaps you can do the time and height together and knock of parts b and c simultaneously! The usual procedure for this sort of thing is to write physics formulas or equations for both motions - the fall and the rise. Ask yourself what kind of motion is going on in each case, then look in your notes or on your formula sheet for distance - time formulas. There may be several of them; pick the one(s) that make use of the information you have and don't require more than one bit of information you don't have. Once you have two good distance formulas for the two motions, I think you will hit upon a way to put them together to find the distance and/or the time of the collision.

 

[nns91]

That's my problem right there. I can't find out what formula to use.

 

[Delphi51]

nns, I'm probably in a different country using different textbooks. Can you list ALL the distance formulas you have for this kind of motion?

 

[nns91]

x= vi*t +at^2/2

x= (1/2)(vi+vf)t

x= vf*t-at^2/2

 

[Delphi51]

Excellent - those look familiar and you know it is accelerated motion!
For the falling ball, you know "vi" and "a" but do not know "vf", so it should be easy to pick the best of the three (must include as many knowns as possible and as few unknowns as possible).

I'm puzzled by how the size of the balls will affect the problem. What exactly is at height 5 m ? Maybe the bottom of the ball so it actually falls 5 m.

 

[nns91]

So I kinda moved on.

I set up 2 equations of position as functions of time.

x1(t) = 10t+ at^2/2

x2(t)= 5(t-1)^2

How can I calculate the acceleration of the ball after bouncing ??

 

[scienture]

I think you've already known that since the 'Perfect elastic collision',the velosity of the ball after bouncing will be as same as that of it before bouncing,which just has an opposite direction.
Now let's try this. Following Newton second law, acceleration comes from force.No matter whether it's before or after bouncing,the only force that put on the ball is its gravity.This doesn't change unless the ball flies to the outer space.
So,since you know the acceleration before bouncing is g=10m/s^2,what is it do you think after bouncing?
You've known the answer now.