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

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$$AOB = Angle between the incident and deflected paths of the ball = $45\degree$
$\angle AOP$ = $\angle BOP$ = $22.5\degree$ = $θ$
Initial and final velocities of the ball = $v$
Horizontal component of the initial velocity = $v\cos \theta$ along $RO$
Vertical component of the initial velocity = $v\sin \theta$ along $PO$
Horizontal component of the final velocity = $v\cos \theta$ along $OS$
Vertical component of the final velocity = $v\sin \theta$ 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
= $mv \cos \theta - (-\;mv \cos \theta)$
= $2mv \cos \theta$
Mass of the ball, $m$ = $0.15 \;kg$
Velocity of the ball, $v$ = $54\; km/h$ = $15 \;m/s$
$\therefore$ Impulse = $2 × 0.15 × 15 \cos 22.5\degree$
= $4.16 \;kg \;m/s$