A batsman deflects a ball by an angle of 90 degree without changing its initial speed , which is equal to 54km/HR .What is the impulse imparted to the ball?(please provide detailed solution with diagram please)
Answers
Given,
Mass of ball = 0.15 Kg
Initial speed of ball =54km/h
= 54×5/18 m/s = 15 m/s
Let the ball inclined along path AO . By batsman deflecting path of it along OB.
Here,
Angle AOP = angle POB = 45/2 °
We have to require break speed of it in components .
We see that horizontal component of velocity ( usin22.5°) remains unchange.
Vertical component (ucos22.5°) just reversed.
Hence,
Change in momentum of the ball = m(v-u)
= mucos22.5° -(-mucos22.5°)
= 2mucos22.5°
= 2 × 0.15 × 15×cos22.5
= 4.16 kgm/s
Hence, impulse imparted by ball = Change in linear momentum of the ball = 4.16 kgm/s
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Given,
Mass of the ball m=0.15kg.
The initial speed of the ball u=54km/h=
60×60
54×1000
=15m/s
The final speed of the ball v
f
=ucos45
0
The momentum change takes place in the ball in the vertical direction only because the initial and final momentum in the horizontal direction is canceled out.
v
i
=−ucos45
0
According to the impulse-momentum theorem,
Impulse=Changeinmomentum
⇒I=m(v
f
−v
i
)
I=mucos45
0
−(−mucos45
0
)
I=2mucos45
0
Substitute the value of mass and velocity.
I=2×0.15×15×0.7071
I=3.19kgm/s