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? (Mass of the ball is 0.15 kg).

Answer 1

Hii dear,

# 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.

Hope that is useful...

Answer 2

Answer:

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

Explanation: