A football of mass 0.5 kg coming towards player with a velocity of 10 m/s. Player hit straight back the football with velocity 15 m/s. If the contact time is 0.02 second then the average force involved is: - (1) 250 N (2) 1250 N (3) 500 N (4) 625 N
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Answer:
GIVEN:-
Mass of the football (m) = 0.5 kg
Initial velocity (u) = 10 m/s
Final velocity (v) = 15 m/s
Contact time (t) = 0.02 sec
TO FIND:-
The average force involved (F)
EXPLANATION:-
From newton's 2nd law ,
>f=m(v-u)/t
>F = 0.5(-25)/0.02
> F = -12.5/0.02
> F = -1250/2
> F = -625 N
Negative sign means retarding force .
EXTRA INFORMATION:-
Derivation of F-ma
The rate of change of momentum of a body is directly proportional to applied force on it :-
Rate of change of momentum=∝ΔP/t
ΔP=change in momentum
t=time
F=force
F∝Pf-Pi/t
F∝mv-mu/t
F∝m(v-u/t)
F∝ma [(v-u/t)=a]
F=kma
K=1 (experimentally)
F=MA
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