#### 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.)

#### Solution 1

The given situation can be represented as shown in the following figure.

Where,

AO = Incident path of the ball

OB = Path followed by the ball after deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = *θ*

Initial and final velocities of the ball =* v*

Horizontal component of the initial velocity = *v*cos *θ* along RO

Vertical component of the initial velocity = *v*sin *θ* along PO

Horizontal component of the final velocity = *v*cos *θ* along OS

Vertical component of the final velocity = *v*sin *θ *along OP

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

∴Impulse imparted to the ball = Change in the linear momentum of the ball

`= mvcostheta - (- mvcos theta)`

'= 2mvcos theta`

Mass of the ball, *m* = 0.15 kg

Velocity of the ball, *v* = 54 km/h = 15 m/s

∴Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

#### Solution 2

Suppose the point O as the position of bat. AO line shows the path along which the ball stikes the bat with velocity u and OB is the path showing deflection such that `angleAOB = 45^@`. Now initial momentum of ball

`= mu cos theta`

= `(0.15 xx 54 xx 1000 xx 22.5)/3600`

= 0.15 xx 15 xx 0.9239 along NO

Final momentum of ball = `mu cos theta` along ON

Impulse = change in momentum = mu cos `theta -(-"mu" cos theta)`

= 2 mu cos theta = 2xx0.15 xx 15 xx 0.9239 = 4.16 kg `m^(-1)`