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PEToronto
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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
V1i = 4.1 m/s
V2i = 1 m/s
V1f = ?
V2f = ?
Homework 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
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
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