Question

A batsman deflect a ball by an angle of 45 without changing its initial speed which is equal to 54km per hour. What is the impulse imparted to the ball mass of the ball is 0.15 kg​

Answer 1

Given,

Mass of ball = 0.15 Kg

Initial speed of ball =54km/h

= 54×5/18 m/s = 15 m/s

Let the ball inclined along path AO . By batsman deflecting path of it along OB.

Here,

Angle AOP = angle POB = 45/2 °

We have to require break speed of it in components .

We see that horizontal component of velocity ( usin22.5°) remains unchange.

Vertical component (ucos22.5°) just reversed.

Hence,

Change in momentum of the ball = m(v-u)

= mucos22.5° -(-mucos22.5°)

= 2mucos22.5°

= 2 × 0.15 × 15×cos22.5

= 4.16 kgm/s

Hence, impulse imparted by ball = Change in linear momentum of the ball = 4.16 kgm/s

Answer 2

Answer:

Explanation:

Answer- 4.16 kgm/s

## Explaination-

# Given-

m = 0.15 kg

v = 54 km/h = 15 m/s

θ = 45°

# Solution-

Assume, vector A be the angle bisector of the ball path such that it makes angle of θ/2 with both initial & final path.

Now, initial momentum is given by,

pi = m×u×cos(θ/2)

Also, final momentum is given by,

pf = -m×u×cos(θ/2)

Impulse = Change in momentum

I = pf-pi

I = m×u×cos(θ/2) + m×u×cos(θ/2)

I = 2×m×u×cos(θ/2)

Putting values,

I = 2×0.15×15×cos22.5°

I = 4.157 kgm/s

Impulse imparted to the ball is 4.157 kgm/s