A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.00 m/s collides head-on with a 0.300 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
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this special scenario, of a head on elastic collision when one of the balls is at rest generates simple analytic solutions
I will give you an outline of the derivation:
call v0 the speed of the first ball before collision, v1 and v2 the speeds of the first and second balls after collision
m1, m2 are the masses of the first and second balls
start with momentum conservation:
m1 v0 = m1 v1 + m2 v2
rewrite as m2 v2 = m1(v0-v1)
energy conservation gives you
1/2 m 1 v0^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2
divide through by 1/2 and rewrite:
m2 v2^2 = m1(v0^2-v1^2) = m1 (v0-v1)(v0+v1)
but from momentum we know that m2 v2 = m1(v0-v1), so substitute this into the energy equation and get
m2 v2^2 = m2 v2 (v0+v1) or v2 = v0+v1
substitute this back into the momentum equation and get that
v1 = (m1-m2)v0/(m1+m2)
which then yields
v2 = 2 m1 v0/(m1+m2)
here, we have
v1 = (0.44kg - 0.3kg)*3m/s / (0.74kg)
v2 = 2*0.44kg x 3m/s /(0.74kg)
both balls are moving in the +x direction
I will give you an outline of the derivation:
call v0 the speed of the first ball before collision, v1 and v2 the speeds of the first and second balls after collision
m1, m2 are the masses of the first and second balls
start with momentum conservation:
m1 v0 = m1 v1 + m2 v2
rewrite as m2 v2 = m1(v0-v1)
energy conservation gives you
1/2 m 1 v0^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2
divide through by 1/2 and rewrite:
m2 v2^2 = m1(v0^2-v1^2) = m1 (v0-v1)(v0+v1)
but from momentum we know that m2 v2 = m1(v0-v1), so substitute this into the energy equation and get
m2 v2^2 = m2 v2 (v0+v1) or v2 = v0+v1
substitute this back into the momentum equation and get that
v1 = (m1-m2)v0/(m1+m2)
which then yields
v2 = 2 m1 v0/(m1+m2)
here, we have
v1 = (0.44kg - 0.3kg)*3m/s / (0.74kg)
v2 = 2*0.44kg x 3m/s /(0.74kg)
both balls are moving in the +x direction