# A cricket ball of mass 150g is moving with a velocity of 12m/s and is hit by a bat, so that the ball is turned back with a velocity of 20m/s. The force of the blow acts for 0.01s on the ball. Find the average force exerted by the bat on the ball. Mass of the ball, m = 150 g = 0.15 kg

Initial speed of the ball, u = -12 m/s

Final speed of the bal after being hit, v = 20 m/s

Impulse = change in momentum = final momentum initial momentum

= Impulse = 0.15 × 20 0.15 × (-12) = 4.8 Ns

• 74 hope helps!

• -24
Force = ma
So, 150*20-150*12
=1200N
• -27
tan(x+y)=tan x+tan y/1-tanx.tany
sin(x+y)/cos(x+y)=sin
• -22
formula for time of flight,maximum height reached and horizontal range
• -24
time taken to reach maximum height

vy=uy+at
vy=o
uy=usin(tita)
vy=uy+at
0=usin(tit)+(-g)t(because acceleration due to gravity is acting downwards (i.e a= -g))
0=usin(tit)-gt
-usin(tita)= -gt
usin(tita)=gt       (- and - gets cancelled)
t=usin(tita)/g

• -18

Mass of the ball, m = 150 g = 0.15 kg

Initial speed of the ball, u = -12 m/s

Final speed of the ball after being hit, v = 20 m/s

Impulse = change in momentum = final momentum initial momentum

= Impulse = 0.15 × 20 0.15 × (-12) = 4.8 Ns

• 13
Here sign convention is used so that one velocity will be positive and other will be negative • 49
12*10^-3
• -7
1200N
• -5
Mass of the ball, m = 150 g = 0.15 kg Initial speed of the ball, u = -12 m/s Final speed of the bal after being hit, v = 20 m/s Impulse = change in momentum = final momentum initial momentum = Impulse = 0.15 × 20 0.15 × (-12) = 4.8 Ns
• -3 • 4
480N
• 2
Mass of the ball, m = 150 g = 0.15 kg

Initial speed of the ball, u = -12 m/s

Final speed of the ball after being hit, v = 20 m/s

Impulse = change in momentum = final momentum initial momentum

= Impulse = 0.15 ? 20 0.15 ? (-12) = 4.8 Ns
• 2 • 1
Mass of ball=150\1000=0.15kg

Velocity of ball =12m\s

Momentum of ball =0.15kg *12m\s =1.8kg m\s

Velocity after collision =20m\s

Momentum of ball =0.15*20 =3kg m\s
Time of contact =0.01s

Av.force = change in momentum ime of contact

= 3 -1.8\0.01s =120 newton
• 0 • -1 • 0
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