Question
A batsman deflects a ball by an angle of $45^o$ without changing its initial speed which is equal to $54\; km/h$. What is the impulse imparted to the ball in $kg\;m/s$? (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
$\angle AOB =$ Angle between the incident and deflected paths of the ball $=45^{\circ}$
$\angle AOP =\angle BOP =22.5^{\circ}=\theta$
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 $=m v \cos \theta-(-m v \cos \theta)$ $=2 m v \cos \theta$
Mass of the ball, $m=0.15\, kg$
Velocity of the ball, $v=54 \,km / h =15 \,m / s$
$\therefore \text { Impulse }=2 \times 0.15 \times 15 \cos 22.5^{\circ}=4.16\, kg\, m / s$
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