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A child bounces a 46 g superball on the side-walk. The velocity change of the superball isfrom 28 m/s downward to 16 m/s upward...Asked by famin
A child bounces a 48 g superball on the sidewalk.
The velocity change of the superball is
from 22 m/s downward to 19 m/s upward.
If the contact time with the sidewalk is 1
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Answer in units of N
The velocity change of the superball is
from 22 m/s downward to 19 m/s upward.
If the contact time with the sidewalk is 1
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Answer in units of N
Answers
Answered by
Henry2,
t = (1/800)s.
V = Vo + a*t = 19.
22 + a*(1/800) = 19,
a/800 = -3,
a = -2400 m/s^2.
F = M*a = 0.048 * (-2400) = -115.2 N.
The negative sign means the force opposes the motion.
V = Vo + a*t = 19.
22 + a*(1/800) = 19,
a/800 = -3,
a = -2400 m/s^2.
F = M*a = 0.048 * (-2400) = -115.2 N.
The negative sign means the force opposes the motion.
Answered by
Henry2,
Correction: V = Vo + a*t = 19 - (-22).
-22 + a*(1/800) = 41.
a/800 = 63,
a = 50,400 m/s^2.
F = M*a = 0.048 * (50,400) = 2419 N.
-22 + a*(1/800) = 41.
a/800 = 63,
a = 50,400 m/s^2.
F = M*a = 0.048 * (50,400) = 2419 N.