Question

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by
the batsman. The ball moves straight back to the bowler after hitting the bat. Assume that the ball rebounds with same
speed and they remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place
would be​

Answer 1

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by  the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that the ball rebounds with same  speed and they remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place  would be​ $1.05 * 10^4 N$.

• Given,
• mass = 150 g =0.15 kg
• Speed = 126 km/h = 35 m/s
• t = 0.001 s
• Formula for calculating the force,
• Force = change in momentum / time
• $F = \dfrac{mv-(-mv)}{t}\\\\=\dfrac{2mv}{t}\\\\=\dfrac{2 * 0.15 * 35}{0.001}\\\\=10500N\\\\=1.05 * 10^5 N$

Answer 2

The force that the batsman had to apply to hold the bat firmly is -1.05 × 10⁴ N

Given:

m = 150 g  = 150/1000 kg = 3/20 kg

u = 126 km/h  = 126 × 1000/(60×60) m/s  = 35 m/s

t = 0.001 s

Explanation:

On assuming the ball rebounds with the same speed.

Thus, the rebound velocity is:

v = - 35 m/s

As the formula of impulse is given as:

F × t = Impulse = Change in momentum of ball

On substituting the values, we get,

F × 0.001 = m(v-u)

F × 0.001 = 3/20 (-35 -35)

F × 0.001 = -21/2

F = -21/(2  × 0.001)

∴ F = -1.05 × 10⁴ N

The negative sign indicates the force is opposite direction to the motion of the ball.