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
Create a free account to view solution
View Solution FreeMore Laws of Motion Questions
A boy of mass 40 kg is hanging from the horizontal branch of a tree. The tension in his arms is minimum when the angle b...A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the for...Two masses m1 and m2 (m1 2) are released from rest from a finite distance. They start under their mutual gravitational a...The length of a spring is α when a force of 4 N is applied on it and the length is β when 5 N force is applied...The breaking strength of a rope is 45 kg wt. A fire man weighing 60 kg slides down the rope. The rope :...