IIT JEE , AIEEE Forum - Physics , Mechanics

rakesh61

A 150 g cricket ball moving horizontally at 20 m/sec was hit straight back to bowler at 12 m/sec if contact with bat lasted for 1/25 sec then average forc eexerted by bat on ball is

Ans 120 N

i want a detailed solution i promise to rate everyone who help me

elessar_iitkgp

The change in momentum of the ball

p = m(v'-v)
Where v' is the final velocity and v is the initial velocity
If

t be the time of contact, then the average force exerted is
F_av =

p/

t
Substitute
m = 0.15 kg
v' = 12 m/s
v = -20 m/s

t = 1/25 s
to get the answer

biki

We know, Force = rate of change of momentum

= m(v-u)/t

0.15(12 + 20)

= --------------------- { taking v as positive and u as negative}

(1/25)

= 12000 N...

CHAMP007

change in momentum=.15 (20+12)=4.8

force=4.8/ 1/25=120N

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priyesh

Let the force exerted be F

now impulse = F * (1/25)sec = change in momentum = 0.15(20) - 0.15(-12) = 0.15 * 32 = 4.8 kgm/sec

therefore F = 4.8 / (1/25) = 4.8 * 25 = 120N

paridhi_aggarwal

see
the impulse is = change in momentum= pf-pi
in this case
pf=0.15*12
pi= -0.15*20
hence, change in momentum= 0.15 [20+12]=4.8
now we know
impulse=ft
therefore
equating the two
F=4.8/ t=120N

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