A snooker ball of mass 0.20kg travelling at a velocity of 14.0ms^-1 hits the side cushion at right angles and rebounds with half the original speed. Calculate:
a) The change in momentum of the ball.
b) The force exerted by the cushion on the ball if it was in contact with the cushion for 0.6s.
a) The change in momentum of the ball.
b) The force exerted by the cushion on the ball if it was in contact with the cushion for 0.6s.
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a) momentum (p) is equal to a mass (m) time a velocity (v). The original momentum will be the mass times the original speed of the ball. The final momentum will be the mass of the ball times one half of the original speed with these two values you can calculate the change in momentum.
b) Force is equal to mass times acceleration (F=ma) and acceleration is change in velocity over change in time (dV/dt). The ball went from 0m/s to half of the original speed in 0.6s (that is your acceleration) and you already know the mass.
b) Force is equal to mass times acceleration (F=ma) and acceleration is change in velocity over change in time (dV/dt). The ball went from 0m/s to half of the original speed in 0.6s (that is your acceleration) and you already know the mass.
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watch out for direction!..momentum is vectr quty!..ie pos/neg..init mom=m*v=0.2*14=2.8kgm/s..final momentum=-0.2*7=-1.4kgm/s..so change in mom=2.8--(-(1.4))=4.2kgm/s
F=D(mv)/dt=4.2/0.6=7N..suspect contact time is prolly more like 0.006=6ms..in which case F=700N
F=D(mv)/dt=4.2/0.6=7N..suspect contact time is prolly more like 0.006=6ms..in which case F=700N
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a) mu - mv (u=original vel , v = final vel)
.2 x 14 - (-.2 x 7)
(final v is neg as it is in opposite direction
change = 4.2 Ns
b) impulse (F x t) = change in momentum
F x .6 = 4.2
F= 7 N
.2 x 14 - (-.2 x 7)
(final v is neg as it is in opposite direction
change = 4.2 Ns
b) impulse (F x t) = change in momentum
F x .6 = 4.2
F= 7 N
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i think b