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A 46g marble moving at 2.3m/s strikes a 24g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on.
a) What is the speed of the first marble immediately after the collision?
b) What is the speed of the second marble immediately after the collision?
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Added Wed, 08 Jul '15
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head on elastic collisions with one particle initially at rest allow us to find analytic solutions
in this collision, both momentum and energy are conserved
momentum conservation gives us
m1 v0 = m1 v1 + m2 v2

m1, m2 are the masses of the objects; v0 is the speed of m1 before collision; v1 and v2 are the speeds of m1 and m2 after collision
energy conservation gives us
1/2 m1 v0^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2

now we can write the momentum equation as
m1 v0 - m1 v1 = m2 v2 or m1(v0-v1) = m2 v2

cancelling out the common factors of 1/2 in the energy equation, we can rearrange that equation as
m1 v0^2 - m1 v1^2 = m2 v2^2

the left hand side becomes m1(v0-v1)(v0+v1), so
m1 (v0-v1)(v0+v1) = m2 v2^2

we have from the momentum equation that m2 v2 = m1 (v0-v1), making this substitution gives us
m2 v2 (v0+v1) = m2 v2^2
v2 = v0+v1

substitute this result for v2 into the momentum equation:
m1(v0-v1) = m2(v0+v1)

write v1 in terms of v0:
v1 = v0(m1-m2)/(m1+m2)

so if masses are in grams and speeds are in m/s, we have
v1 = 2.3m/s(46g-24g)/(46g+24g) = 0.72285m/s

since v2=v0+v1, v2 = 2.3m/s + 0.72285m/s = 3.02285m/s
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Added Wed, 08 Jul '15
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