A cricket ball of mass 150 g moving at a speed of 25 m s^-1 is brought to rest by a player in 0.03 s. Find the average force applied by the player.

The 1^st equation of motion states that;

v = u + at

Given,

m = 150 g

Converting g to kg,

1000 g = 1 kg

150 g = ($\dfrac{1}{1000}$) x 150

150 g = 0.15 kg

Hence, m = 0.15 kg

u = 25 m s^-1

v = 0

t = 0.03 s

Substituting the values in the formula above we get,

$0 = 25 + (a \times 0.03) \\[0.5em] \Rightarrow a = -\dfrac{25}{0.03} \\[0.5em] \Rightarrow a = - 833.33 \\[0.5em]$

Hence, acceleration of the ball = - 833.33 m s^-2`

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$F = 0.15 \times (- 833.33) \\[0.5em] \Rightarrow F = -125 N \\[0.5em]$

Hence,

Magnitude of force = -125 N