# Two objects colliding and dropping at different distances

1. Jun 9, 2012

### thisischris

A ball bearing is rolled down a ramp clamped at the edge of a bench. The ball bearing makes a head-on collosion with a marble at the bottom of the ramp.

My problem lies with question B.

A: Initially the ball bearing rolls down without a marble in place. The ball bearing falls a vertical distance of 1.2m and lands a horizontal distance of 0.95m from the end of the ramp. Show speed is 2ms.

Used s = ut + (0.5)t2 to calculate the time in 'air' = 0.49s.
Used s = ut + 0.5at again:
(s - 0.5at2)/t = u
(0.95 - 0) / 0.49 = 1.94 ms-1

B: When the ball bearing is rolled down the ramp with a marble in place, it knocks the marble forwards. The ball bearing lands a horizontal distance of 0.64m from ramp. The marble lands a further 0.93m from ramp. Calculate the ratio of masses of the ball bearing and the marble.

Answer: Momentum conservation, correct substitutions, m1 : m2 = 5.3

My attempt:
m1u + m2u = m1v + m2v
m1u = m1v + m2v As marble has 0 momentum initally.
m1(u - v) =m2v
m1(1.94 - v) =m2v

I then calculated m1's final velocity by using speed = distance / time. Hence v = 0.64 / 0.49 = 1.31ms-1
m2's final velocity by using speed = distance / time. Hence v = 0.64 + 0.93 / 0.49 = 3.20ms-1

m1(1.94 - 1.31) =m2(3.2)
m1(0.63) =m2(3.20)
m1 = m2(5.08)

I can't seem to understand why I'm quite a bit out.

Thank you :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 9, 2012

### LawrenceC

The 1.94 is off by a little. Carry more significant figures in intermediate calculations, then round off at end of computation. Problem is caused by your computation in part A for time.