Question 5.20 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.)
Chapter Laws Of Motion Page 111
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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
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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