# SOLVED: Elastic Collision, Same Direction

1. Nov 30, 2014

### PEToronto

1. The problem statement, all variables and given/known data
A 100g ball moving to the right at 4.1m/s catches up and collides with a 400g ball that is moving to the right at 1.0m/s .

a)
If the collision is perfectly elastic, what is the speed and direction of the 100g ball after the collision?

b)
If the collision is perfectly elastic, what is the speed and direction of the 400g ball after the collision?

Given Var.
m1 = 100 g = 0.1 kg
m2 = 400 g = 0.4 kg

V1i = 4.1 m/s
V2i = 1 m/s

V1f = ?
V2f = ?

2. Relevant equations
pi = pf
p1i + p2i = p1i + p2i
m1V1i + m2V2i = m1V1f + m2V2f

Ki = Kf
K1i + K2i = K1i + K2i
(1/2)m1V1i2 + (1/2)m2V2i2 = (1/2)m1V1f2 + (1/2)m2V2f2

3. The attempt at a solution

I first set up the equation for conservation of momentum:
pi = pf
p1i + p2i = p1i + p2i
m1V1i + m2V2i = m1V1f + m2V2f
(0.1)(4.1) + (0.4)(1) = 0.1V1f + 0.4V2f
2.081 = 0.1V1f + 0.4V2f

Solving for V1f
V1f = (0.81 - 0.4V2f) / 0.1

Now the equation for Conservation of energy:
Ki = Kf
K1i + K2i = K1i + K2i
(1/2)m1V1i2 + (1/2)m2V2i2 = (1/2)m1V1f2 + (1/2)m2V2f2
(0.1)(4.1)2 + (0.4)(1)2 = 0.1V1f2 + 0.4V2f2

Subbing the previous identity of V1f:
2.081 = (0.1)(4.1)2 + (0.4)(1)2 = 0.1((0.81 - 0.4V2f) / 0.1)2 + 0.4V2f2
2.081 = 6.561 - 6.48V2f + 1.6V2f2 + 0.4V2f2

Solving for V2f gives 0.9999 and 2.24
2.24 m/s is more reasonable, as the ball must have gained velocity during the collision rather than lost it.

Now solving for V1f:
2.081 = 0.1V1f + 0.4V2f
2.081 = 0.1V1f + 0.4(2.24)
V1f = (2.081 - 0.896) / 0.1
V1f = 11.85 m/s

I am concerned because I did not get a negative value for V1f, which I was expecting because it seems like the smaller first ball will rebound in the opposite direction. Is it also concerning that the final velocity of the smaller ball is much higher than the initial? am I missing a minus sign somewhere?

Thank you very much for taking a look

Last edited by a moderator: Nov 30, 2014
2. Nov 30, 2014

### PEToronto

By plugging V2f into the Kinetic Energy Equation and solving for V1f, I get +/- 0.859. -0.859 seems more reasonable. Did I mess up again?

3. Nov 30, 2014

### haruspex

Looks right.
From momentum and energy conservation one can deduce this general equation for the components of velocity along the direction of impact: v1i-v2i = v2f-v1f. This Newton's Experimental Law for the special case of perfectly elastic collision. This often helps avoid getting tangled up in solving quadratics, producing a spurious extra solution.
As an exercise, see if you can derive it.

4. Nov 30, 2014

### PEToronto

Thanks haruspex,
That formula looks very useful, and I'll try deriving it when i finish my homework.
V1f turned out to be -0.86 m/s. I'll rename this thread as 'solved'.

Thanks again!
PET