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 = \(2 × 0.15 × 15 \cos 22.5\degree\) 
= \(4.16 \;kg \;m/s\)