Question

A cricket ball of mass 100g moving with a speed of 20 ms⁻¹ is brought to rest by
a player in 0.02 s. Find
i) the change in momentum of ball.
ii) the average force applied by the ball.

Answer 1

Explanation,

Given,

• Mass (m) of a cricket ball is 100g.
• Initial velocity (u) = 20 ms-¹
• Final velocity (v) = 0 ms-¹
• Time (t) = 0.02 s

To Find,

1. The change in momentum of ball.
2. The average force applied by the ball.

Solution,

Firstly, convert gram into kg.

⇒ Mass (m) = 100 g

⇒ m = 100/1000

⇒ m = 0.1 kg

Case (I), The change in momentum of ball.

[ For initial momentum ]

Pi = mu

Here,

• Pi = Initial momentum
• m = Mass
• u = Initial velocity

[ Put the values ]

⇒ Pi = 0.1 × 20

⇒ Pi = 2 kg. m/s

[ For final momentum ]

Pf = mv

Here,

• Pf = Final momentum
• m = Mass
• v = Final velocity

[ Put the values ]

⇒ Pf = 0.1 × 0

⇒ Pf = 0 kg. m/s

So,

Δp = Pf - Pi

Where,

• Δp = Change in momentum
• Pf = Final momentum
• Pi = Initial momentum

[ Put the values ]

⇒ Δp = 0 - 2

⇒ Δp = -2 kg.m/s

Case (II), The average force applied by the ball.

Applying newton's 1st equation of motion,

v = u + at

Here,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time

[ Put the values ]

⇒ 20 = 0 + a × 0.02

⇒ 20 = 0.02a

⇒ a = 20/0.02

⇒ a = 1000 m/s²

According to the newton's 2nd law of motion,

F = ma

Here,

• F = Force
• m = Mass
• a = Acceleration

[ Put the values ]

⇒ F = 0.1 × 1000

⇒ F = 100N

Answer 2

Answer: Mass= 0.1 kg

Initial Momentum= m*speed (u)=0.1*30= 3

Final Momentum= m* rest (v)= 0.1*0 = 0

Change in Momentum= Final-Intial = 0-3 = -3

Rate of change in momentum= (change in momentum)÷time taken  

                                              = (-3)÷0.03

                                              = -100