Laws of MotionHard
Question
A ball weighing 10 gm hits a hard surface vertically with a speed of 5m/s and rebounds with the same speed. The ball remain in contact with the surface for 0.01 sec. The average force exerted by the surface on the ball is :
Options
A.100 N
B.10 N
C.1 N
D.150 N
Solution
P = 2mv
= 2 × 10 × 10-3 × 5
= 10-1
ᐃt = 10-2
F =
=
= 10 N
= 2 × 10 × 10-3 × 5
= 10-1
ᐃt = 10-2
F =
=
= 10 N
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