SOLVED: Elastic Collision, Same Direction

[PEToronto]

Homework Statement

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

V_1i = 4.1 m/s
V_2i = 1 m/s

V_1f = ?
V_2f = ?

Homework Equations

p_i = p_f
p_1i + p_2i = p_1i + p_2i
m₁V_1i + m₂V_2i = m₁V_1f + m₂V_2f

K_i = K_f
K_1i + K_2i = K_1i + K_2i
(1/2)m₁V_1i² + (1/2)m₂V_2i² = (1/2)m₁V_1f² + (1/2)m₂V_2f²

The Attempt at a Solution

I first set up the equation for conservation of momentum:
p_i = p_f
p_1i + p_2i = p_1i + p_2i
m₁V_1i + m₂V_2i = m₁V_1f + m₂V_2f
(0.1)(4.1) + (0.4)(1) = 0.1V_1f + 0.4V_2f
2.081 = 0.1V_1f + 0.4V_2f

Solving for V_1f
V_1f = (0.81 - 0.4V_2f) / 0.1

Now the equation for Conservation of energy:
K_i = K_f
K_1i + K_2i = K_1i + K_2i
(1/2)m₁V_1i² + (1/2)m₂V_2i² = (1/2)m₁V_1f² + (1/2)m₂V_2f²
(0.1)(4.1)² + (0.4)(1)² = 0.1V_1f² + 0.4V_2f²

Subbing the previous identity of V_1f:
2.081 = (0.1)(4.1)² + (0.4)(1)² = 0.1((0.81 - 0.4V_2f) / 0.1)² + 0.4V_2f²
2.081 = 6.561 - 6.48V_2f + 1.6V_2f² + 0.4V_2f²

Solving for V_2f 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 V_1f:
2.081 = 0.1V_1f + 0.4V_2f
2.081 = 0.1V_1f + 0.4(2.24)
V_1f = (2.081 - 0.896) / 0.1
V_1f = 11.85 m/s

I am concerned because I did not get a negative value for V_1f, 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

 

[PEToronto]

By plugging V_2f into the Kinetic Energy Equation and solving for V_1f, I get +/- 0.859. -0.859 seems more reasonable. Did I mess up again?

 

[haruspex]

By plugging V_2f into the Kinetic Energy Equation and solving for V_1f, I get +/- 0.859. -0.859 seems more reasonable. Did I mess up again?

Looks right.
From momentum and energy conservation one can deduce this general equation for the components of velocity along the direction of impact: v_1i-v_2i = v_2f-v_1f. 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.

 

[PEToronto]

Thanks haruspex,
That formula looks very useful, and I'll try deriving it when i finish my homework.
V_1f turned out to be -0.86 m/s. I'll rename this thread as 'solved'.

Thanks again!
PET