Question

A batsman hits back a ball straight in the direction of the bowler without changing its speed of 10 m/s. The mass is 0.15 kg. (a) What impulse is imparted to the ball? (b) Assuming that the ball remained in contact with the bat for 0.001 s, what is the average force exerted by the bat on the ball?​

Answer 1

Answer:

(A) Impulse imparted = -3 kg m/s

(B) Force exerted by bat = 3000 N

Explanation:

$\blacksquare$ We know that,

$\implies$ Impulse is defined as the change in momentum

$\implies$ Impulse = Δp

$\implies$ Impulse = Final momentum - Inital momentum

$\implies$ Impulse = mv - mu    --------(1)

$\blacksquare$ We also know that,

$\implies$ Force is defined as the rate of change of momentum

$\implies$ Force = $\dfrac{\Delta p}{\Delta t}$

$\implies$ Force = $\dfrac{Change \ in \ momentum}{Change \ in \ time}$  --------(2)

$\blacksquare$ Given that,

$\diamond$ Initial velocity = u = 10 m/s

$\diamond$ Final velocity = v = -10 m/s

(-ve sign indicates that ball is moving in opposite direction after it is hit by the bat)

$\diamond$ Mass = m = 0.15 kg

$\diamond$ Change in time = Δt = 0.001 s

$\blacksquare$ Substituting these values in (1) we get,

$\implies$ Impulse = (0.15 * -10) - (0.15 * 10)

$\implies$ Impulse = -1.5 - (1.5)

$\implies$ Impulse = -1.5 - 1.5

$\implies$ Impulse = -3 kg m/s

$\blacksquare$ Substituting the values in (2) we get,

$\implies$ Force = $\dfrac{\Delta p}{\Delta t}$

$\implies$ Force = $\dfrac{-3}{0.001} =-3000 \ N$

Answer 2

Explanation: Change in momentum +0.15×12−(−0.15×12)=3.6 N s

Impulse = 3.6 N s in the direction from the batsman to the bowler.