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Finding time and momentum after collision?

  1. Nov 23, 2008 #1
    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.

    a) How long does it take the heavy ball to hit the floor after the collision?
    I tried solving it this way:
    (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
    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=(.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

    = (.00325kg)(.815 m/s)
    = .00265 kg*m/s <----this was my answer.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 23, 2008 #2


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    Homework Helper

    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?
    Last edited: Nov 23, 2008
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