Question

A ball weighing 10 ghits 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) sec. The average force exerted by the
surface on the ball is :-
(1) 100 N
(2) 10 N
(3) 1 N
(4) 0.1 N
A force-time graph for the motion of a body is​

Answer 1

Answer:

b

Explanation:

Here, m=10g=101000kg=0.01kg

υ1=−5m/s,υ2=+5m/s,t=0.01sec

According to impulse momentum theorem

Impulse = change in momentum

Fav×t=mυ2−m(−υ1)=mυ2+mυ1

Fav×0.01=0.01×5+0.01×5

Fav=0.10.01=10N .

Answer 2

Answer:

1000g =1kg.

1000m =1km

1000ml= 1l.