Question
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg)
✓
Answer
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
$\angle\text{AOB}$ = Angle between the incident and deflected paths of the ball = 45°
$\angle\text{AOP}=\angle\text{BOP}=22.5^\circ=\theta$
Initial and final velocities of the ball = v
Horizontal component of the initial velocity = $\text{v}\cos\theta$ along RO
Vertical component of the initial velocity = $\text{v}\sin\theta$ along PO
Horizontal component of the final velocity = $\text{v}\cos\theta$ along OS
Vertical component of the final velocity = $\text{v}\cos\theta$ along OP
The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.
$\therefore$ Impulse imparted to the ball = Change in the linear momentum of the ball
$=\text{mv}\cos\theta-(-\text{mv}\cos\theta)=2\text{mv}\cos\theta$
Mass of the ball, m = 0.15kg
Velocity of the ball, v = 54km/h = 15m/s
$\therefore$ Impulse $2\times0.15\times15\cos22.5^\circ=4.16\text{kgm/s}$
Where,
AO = Incident path of the ball
OB = Path followed by the ball after deflection
$\angle\text{AOB}$ = Angle between the incident and deflected paths of the ball = 45°
$\angle\text{AOP}=\angle\text{BOP}=22.5^\circ=\theta$
Initial and final velocities of the ball = v
Horizontal component of the initial velocity = $\text{v}\cos\theta$ along RO
Vertical component of the initial velocity = $\text{v}\sin\theta$ along PO
Horizontal component of the final velocity = $\text{v}\cos\theta$ along OS
Vertical component of the final velocity = $\text{v}\cos\theta$ along OP
The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.
$\therefore$ Impulse imparted to the ball = Change in the linear momentum of the ball
$=\text{mv}\cos\theta-(-\text{mv}\cos\theta)=2\text{mv}\cos\theta$
Mass of the ball, m = 0.15kg
Velocity of the ball, v = 54km/h = 15m/s
$\therefore$ Impulse $2\times0.15\times15\cos22.5^\circ=4.16\text{kgm/s}$
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