Question -
Answer -
Thegiven 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
∠AOB = Angle between the incident anddeflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
Initial and final velocities of the ball = v
Horizontal component of the initial velocity= vcos θ along RO
Vertical component of the initial velocity= vsin θ along PO
Horizontal component of the final velocity= vcos θ along OS
Vertical component of the final velocity= vsin θ along OP
Thehorizontal components of velocities suffer no change. The vertical componentsof velocities are in the opposite directions.
∴Impulse imparted to the ball = Change in thelinear momentum of the ball
Mass of the ball, m = 0.15kg
Velocity of the ball, v =54 km/h = 15 m/s
∴Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kgm/s