A batsman deflects a ball by an angle of degree 45 without changing its initial speed which is equal to 54km/hr. what is the impulse imparted to the ball? mass of the ball is 0.15 kg.
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impulse = 2 m v cos (45/2) = change in momentum
= 2 * 0.15 * 54*5/18* cos(22.5) kg-m/s
= 4.15 kg-m/s
= 2 * 0.15 * 54*5/18* cos(22.5) kg-m/s
= 4.15 kg-m/s
Answered by
1
Answer:
Explanation:
Answer- 4.16 kgm/s
## Explaination-
# Given-
m = 0.15 kg
v = 54 km/h = 15 m/s
θ = 45°
# Solution-
Assume, vector A be the angle bisector of the ball path such that it makes angle of θ/2 with both initial & final path.
Now, initial momentum is given by,
pi = m×u×cos(θ/2)
Also, final momentum is given by,
pf = -m×u×cos(θ/2)
Impulse = Change in momentum
I = pf-pi
I = m×u×cos(θ/2) + m×u×cos(θ/2)
I = 2×m×u×cos(θ/2)
Putting values,
I = 2×0.15×15×cos22.5°
I = 4.157 kgm/s
Impulse imparted to the ball is 4.157 kgm/s
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