Asked by Erin
A 9.1 g ball is hit into a 98 g block of clay at rest on a level surface. After impact, the block slides 8.0 m before coming to rest. If the coefficient of friction is 0.60, determine the speed of the ball before impact.
Answers
Answered by
Damon
initial ball speed = vi
momentum before = .0091 vi
momentum after = .0091 vi = (.107)vf
so
vf = .085 vi
work done by friction = initial Ke of clay with ball
.6 (.107)(9.81)(8) = (1/2)(.107)(vf^2)
so vf = 8.86 m/s
vi = 8.86/.085 = 104 m/s
check my arithmetic !
momentum before = .0091 vi
momentum after = .0091 vi = (.107)vf
so
vf = .085 vi
work done by friction = initial Ke of clay with ball
.6 (.107)(9.81)(8) = (1/2)(.107)(vf^2)
so vf = 8.86 m/s
vi = 8.86/.085 = 104 m/s
check my arithmetic !
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