Question

A ball weighing 10 g hits a hard surface
vertically with a speed of 5 m/s and rebounds with
the same speed. The ball remains in contact with
the surface for 0.01 second. What is the average
force exerted by the surface on the ball?
A) 1 N
B) 0.1 N
C) 100 N
D) 10 N

Answer 1

Answer:

1N

Explanation:

F=ma

a= v/t

a= 5/0.01

a=500 m/sec²

F=10/1000×500

F=1N

Answer 2

Answer:

Since, F=P/T

where, P= change in momentum and T = total time

F= ( mv1 - mv2 )/T

F= { 10[5-(-5)]}/0.01

F=1000

F=1000/1000=1N

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