Question

5. A cricket ball is moving with velocity of 10 m/s and is hit by bat so it turns back and moves with velocity of 15m/s. Mass of ball is 150 g . Find force by batsman(bat) if time of interaction between bat and ball is 0.02 second.

Answer 1

Initial velocity (u) = -10 m/s
Final velocity (v) = 15 m/s
Mass of the ball (m) = 150 g = 150/1000 = 0.15 kg
Time of interaction (t) = 0.02 second.

   v = u + at
⇒a = (v - u) / t     [Acceleration is the change in velocity per time taken]
      = (15 -(-10)) / 0.02 
      = (15 + 10) / 0/02
      = 25 / 0.02
      = 1250 m/s²

Force (F) = m × a
              = 0.15 × 1250
              = 187.5 kg.m/s² or 187.5 N

Answer 2

Answer:

A)187.5 N

Explanation:

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I also went to give big bang edge 8th class...

Actually you can solve this question by 2 ways-

1)initial velocity(u)=10 m/sec

Final velocity(v)=-15 m/sec(we take negative value because it's moving opposite to the direction)

Time = 0.02 sec

Acceleration=(v-u)/t

=-15-10/0.02

=-25/0.02

=-1250 m/sec^2

Mass of the ball=150g=150÷1000 kg

=0.15 kg

Force=ma

Force=1250×0.15

=187.5 N, (A)

2)mass of ball=150÷1000 kg=0.15kg

Initial velocity(u)=10 m/sec

Initial momentum=10×0.15=1.5 kg m/sec

Final velocity(v)=-15 m/sec

Final momentum=-15×.15=2.25 Kg m/sec

Impulse=final momentum- initial momentum

Impulse=-2.25-1.5=-3.75

Time=0.02 sec

By newton's second law of motion,

Force=impulse/time

Force=3.75/0.02

=187.5 N