Asked by Student
A 70-g ball B dropped from a height h_0=1.5 m reaches a height h_2=0.25 m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine
(a) the coefficient of restitution between the ball and the plates,
(b) the height h_1 of the ball’s first bounce.
(a) the coefficient of restitution between the ball and the plates,
(b) the height h_1 of the ball’s first bounce.
Answers
Answered by
Damon
I do not care what the mass is, call it m
If it bounces off the same thing twice (I do not get it about the two plates because you provide data for only one of them)
v = velocity at impact with plate A
(1/2) m v^2 = m g h
v = sqrt(2 gh) = sqrt (2*9.81*1.5) = sqrt (29.43) = 5.42 m/s
w = velocity headed up from first hit = 5.42 k
where k is our coef of restitution
we lose no energy going up and then coming down so
we hit plate B at speed w = 5.42 k
we rebound from plate B at speed x = 5.42 k^2
now
m g h = m g (0.25) = (1/2) m (5.42k^2)^2
9.81 (.25) = .5 (29.4) k^4
k^4 = .167
k^2 = .408
k = .639
That should get you started although I am not sure what the two plates is about.
If it bounces off the same thing twice (I do not get it about the two plates because you provide data for only one of them)
v = velocity at impact with plate A
(1/2) m v^2 = m g h
v = sqrt(2 gh) = sqrt (2*9.81*1.5) = sqrt (29.43) = 5.42 m/s
w = velocity headed up from first hit = 5.42 k
where k is our coef of restitution
we lose no energy going up and then coming down so
we hit plate B at speed w = 5.42 k
we rebound from plate B at speed x = 5.42 k^2
now
m g h = m g (0.25) = (1/2) m (5.42k^2)^2
9.81 (.25) = .5 (29.4) k^4
k^4 = .167
k^2 = .408
k = .639
That should get you started although I am not sure what the two plates is about.
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