A spherical ball of mass 1 kg and radius 3 cm is attached to a wire of length 2
ID: 1342781 • Letter: A
Question
A spherical ball of mass 1 kg and radius 3 cm is attached to a wire of length 2 m to form a pendulum. The top of the pendulum is attached to a wall near the top. The wire is pulled straight out horizontally and the ball is released from rest. The ball strikes the wall and rebounds a horizontal distance of 1 m. The wall is dented distance of 1 cm.a) What is the speed of the ball when it strikes the wall?
b) What is the speed of the ball when it leaves the wall?
c) How much energy did the ball lose?
d) What is the average force exerted by the ball on the wall?
e) What is the change in momentum of the ball?
f) How long was the ball in contact with the wall? A spherical ball of mass 1 kg and radius 3 cm is attached to a wire of length 2 m to form a pendulum. The top of the pendulum is attached to a wall near the top. The wire is pulled straight out horizontally and the ball is released from rest. The ball strikes the wall and rebounds a horizontal distance of 1 m. The wall is dented distance of 1 cm.
a) What is the speed of the ball when it strikes the wall?
b) What is the speed of the ball when it leaves the wall?
c) How much energy did the ball lose?
d) What is the average force exerted by the ball on the wall?
e) What is the change in momentum of the ball?
f) How long was the ball in contact with the wall?
a) What is the speed of the ball when it strikes the wall?
b) What is the speed of the ball when it leaves the wall?
c) How much energy did the ball lose?
d) What is the average force exerted by the ball on the wall?
e) What is the change in momentum of the ball?
f) How long was the ball in contact with the wall?
Explanation / Answer
a) Using energy conservation,
mgL = mv^2 /2
v = sqrt(2gL) = sqrt(2 x 9.81 x 2) = 6.26 m/s^2
b) horizontal distance = Lsin@ = 1
@ = sin^-1(1/2) = 30 deg
h = L - Lcos30 =0.268 m
v = sqrt(2 x 9.81 x 0.268) =2.29 m/s
c) energy lost = 1 x 6.26^2 /2 - 1 x2.29^2 /2 = 16.97 J
d) using v^2 - u^2 = 2ad
2.29^2 - 6.26^2 =2 x a x 0.01
a =1697.17 m/s^3
F = ma = 1697.17 N
e) change in momentum = (1 x 6.26) - (1 x 2.92) = 3.34 kg .m/s
f) change in momentum = F x t
t = 1.97 x 10^-3 sec
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