Asked by aparna
A spherical metal ball of mass 0.5kg moving with a speed of 0.5 ms-1on a smooth linear horizontal track collides head on with another ball B of same mass at rest. Assuming the collision to be perfectly elastic, what are the speeds of A and B after collision?
Answers
Answered by
Damon
momentum before = .5 * .5 + .5 * 0 = .25
energy before = (1/2) (.5) .5^2 = (1/2) .5^3/2
momentum after = .5 * v1 + .5 V2 = .25 still
energy after = (1/2) .5 v1^2+(1/2)(.5) v2^2=(1/2)(.5)(v1^2+v2^2)
=(1/2)(.5)^3/2 still
so v1+v2 = .5 so v2=.5-v1
and
v1^2+ v2^2 = .5^1/2
v1^2 + .25 - v1 + v1^2 = .25
2 v1^2 - v1 = 0
v1 (2v1-1) =0
either v1 = 0 or v1 = 0.5 after crashing through ball 2 without losing any speed
v1 = 0 and v2 = .5 -0 = .5
the first one stops, the second one goes on at original speed.
energy before = (1/2) (.5) .5^2 = (1/2) .5^3/2
momentum after = .5 * v1 + .5 V2 = .25 still
energy after = (1/2) .5 v1^2+(1/2)(.5) v2^2=(1/2)(.5)(v1^2+v2^2)
=(1/2)(.5)^3/2 still
so v1+v2 = .5 so v2=.5-v1
and
v1^2+ v2^2 = .5^1/2
v1^2 + .25 - v1 + v1^2 = .25
2 v1^2 - v1 = 0
v1 (2v1-1) =0
either v1 = 0 or v1 = 0.5 after crashing through ball 2 without losing any speed
v1 = 0 and v2 = .5 -0 = .5
the first one stops, the second one goes on at original speed.
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