Asked by Trevor
A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.
A) What is the speed of the block after the collision?
B) What percentage of the ball's initial energy is "lost"?
A) What is the speed of the block after the collision?
B) What percentage of the ball's initial energy is "lost"?
Answers
Answered by
Damon
.050 Vo = 1.050 v
so
v = .0476 Vo
initial Ke = (1/2)(.05)Vo^2
final Ke = (1/2)(1.05)v^2
100 (.05 Vo^2 - 1.05 v^2) /.05 Vo^2
100 (.05 Vo^2 - .00238 Vo^2)/.05 Vo^2
= 95.2 % of initial energy is lost
so
v = .0476 Vo
initial Ke = (1/2)(.05)Vo^2
final Ke = (1/2)(1.05)v^2
100 (.05 Vo^2 - 1.05 v^2) /.05 Vo^2
100 (.05 Vo^2 - .00238 Vo^2)/.05 Vo^2
= 95.2 % of initial energy is lost
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