A batsman deflects a ball by an angle of 45^o | vedclass

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Question

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

Vedclass · Accepted Answer

$4.16$

Vedclass · 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 $=45^{\circ}$

$\angle AOP =\angle BOP =22.5^{\circ}=	heta$

Initial and final velocities of the ball $=v$

Horizontal component of the initial velocity $=v \cos 	heta$ along $RO$

Vertical component of the initial velocity $=v \sin 	heta$ along $PO$

Horizontal component of the final velocity $=v \cos 	heta$ along $OS$

Vertical component of the final velocity $=v \sin 	heta$ along $OP$

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

$	herefore$ Impulse imparted to the ball $=$ Change in the linear momentum of the ball $=m v \cos 	heta-(-m v \cos 	heta)$ $=2 m v \cos 	heta$

Mass of the ball, $m=0.15\, kg$

Velocity of the ball, $v=54 \,km / h =15 \,m / s$

$	herefore 	ext { Impulse }=2 	imes 0.15 	imes 15 \cos 22.5^{\circ}=4.16\, kg\, m / s$

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 $0.15 \;kg.)$

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