Question

a ball of mass 10g is dropped on the ground from a height of 10m. it rebounds to a height of 2.5m . if the Ball is in contact with the ground for 0.01s, then coefficient of restitution between the ground and ball is​

Answer 1

Answer:

Hey mate! here is your answer

Explanation:

Divide it into two parts

First find the value of velocity when it will going to collide with the floor let say it is v1

secondly find out the velocity INITIAL ONE for the second part taking your displacement 2.5m let say it is v2

Now take t=0.02s take initial velocity to be equal to (-v1)

final velocity to be equal to v2 apply the first equation of motion and find the acceleration.

g=9.8m/s^2

u=0

v^2/19.6=10

v=14m/s

while bouncing back

0^2-u^2/-19.6=2.5

u^2=49

u=7m/s

time in contact=0.02s

therefore, (v-u)/t=14-7/0.02=350m/s^2

a=350m/s^2

V1=√2gh1= √2×9.8×10=√196=14

V2=√2gh2=√2×9.8×2.5=√49=7

a=F/m

F=m[v1-(v2)]/t

a=m[v1-(-v2)]/m×0.02=[14+7]/0.02=2100/2=1050