Question

A batsman deflects a ball by an angle of 45 without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? If the time of contact between the ball and the bat is 0.01s, what is the average force exerted by the bat on the ball? (mass of the ball is 0.15 kg.)

Answer 1

The average force exerted by the bat on the ball is 4.16 kg m/sec.

Given-

• Angle of deflection of the ball is = 45⁰
• Initial speed of the ball = 54 km/h
• Time of contact between the ball and the bat = 0.01 sec
• Mass of the ball = 0.15 kg

From the attached figure it is clear that AO is the incident path of the ball and OB is the path followed by the ball after deflection.

∠AOB = 45⁰

∠AOP = ∠BOP = 22.5⁰ = θ

Let the initial and final velocity of the ball be v

From the figure it is clear that

Horizontal component of initial velocity = vcosθ

Vertical component of initial velocity = vsinθ

Horizontal component of final velocity = vcosθ

Vertical component of final velocity = vsinθ

We know that impulse imparted to the ball is equals to the change in linear momentum of the ball.

= mvcosθ-(-mvcosθ) = 2mvcosθ

By substituting the values we get

Impulse = 2 × 0.15 × 15 cos 22.5 = 4.16 kg m/sec