ICSE Class 9 Physics Question 5 of 20

Laws of Motion — Question 10

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Question

Question 10

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.

Answer

The 1st equation of motion states that;

v = u + at

Given,

m = 150 g

Converting g to kg,

1000 g = 1 kg

150 g = (11000\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×0.03)a=250.03a=833.33 m s20 = 25 + (\text a \times 0.03) \\[0.5em] \Rightarrow \text a = -\dfrac{25}{0.03} \\[0.5em] \Rightarrow \text a = - 833.33\ \text {m s}^{-2} \\[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×(833.33)F=125 N\text F = 0.15 \times (- 833.33) \\[0.5em] \Rightarrow \text F = -125\ \text N \\[0.5em]

Hence,

Magnitude of force = -125 N