Finding time and momentum after collision?

[darkchii77]

1. PROBLEM: Students roll a heavy ball (m=5.39g) down a ramp (that ends horizontally) into a light ball (m=3.25g). The top of the ramp is 13.5 cm above the bottom of the ramp. The bottom of the ramp is 1.35 m above the floor. The students found that without collision, the heavy ball travels 89.0 cm away from the table before hitting the floor. After the head-on collision, the heavy ball travels 51.5 cm away from the table before hitting the floor.

RELEVANT EQUATIONS & MY SOLUTIONS:
a) How long does it take the heavy ball to hit the floor after the collision?
I tried solving it this way:
PE=mgh
KE=(1/2)mv^2
PE=KE
(9.8 m/s^2)(.15m)=(1/2)v^2
v=1.63 m/s


I did a Pythagorean triangle to find the distance(hypotenuse) between 1.35m and .515 m and got 1.44 m
distance = rate/time
(1.44m)=(1.63m/s)(t)
t=.88 sec.

I didn't know what i was doing at all so it would be so great if someone could explain how to do this problem. please don't mind my work since i know like 90% of it is wrong and completely doesn't make sense. please please please help! ;_;

b)What is the momentum of the heavy ball before the collision?
p=mv
p=(.00539kg)(1.63m/s)= .00877 kg*m/s

This i understand, but i know i got the velocity wrong from the previous question.

c) What should be the momentum of the light ball after the collision?
v1+v2= -(2vf) <----um...i thought this was inelastic and somehow added the final velocities.

1.63 m/s + 0= -(2vf)
vf=.815 m/s

p=mv
= (.00325kg)(.815 m/s)
= .00265 kg*m/s <----this was my answer.

 

[LowlyPion]

Welcome to PF.

The time to fall is generally given by x = 1/2*g*t²

Since you started with velocity all horizontal after the collision, then your vertical component would be starting from 0 so that should be all you need.

And you are right you calculated your V in a) incorrectly.

In c) the time to fall divided into the distance from the table should give you the velocity of the small ball after the collision shouldn't it?