The figure shows a 2.75 kg softball before and after it hits a bat. Before the c

Question

The figure shows a 2.75 kg softball before and after it hits a bat. Before the collision, the magnitude of the balls velocity (v1) is 15 m/s at an angle of 30 degrees with the horizontal. The batter hits a pop up and the ball travels straight up after it hits the bat with a velocity of magnitude 9 m/s. The collision last for 2.10 ms.

A)Find the magnitude of the impulse on the ball by the bat

B)Find the direction of the impulse

C)Find the magnitude of the average force on the ball

D)Find the direction of the average force

Explanation / Answer

Here ,

initial velocity , u = - 15 *(cos(30) i + j *sin(30))

final velocity , v = 9 j m/s

A)

as impulse = change in momentum

Impulse = maas * ( v - u)

Impulse = 2.75 *(9 j - (- 15 *(cos(30) i + j *sin(30))))

Impulse = 35.7 i +45.4 j N.s

magnitude of impulse = sqrt(35.7^2 + 45.4^2)

magnitude of impulse = 57.7 N.s

the magnitude of impulse is 57.7 N.s

part B)

direction = arctan(45.4/35.7)

direction = 51.8 degree

the direction of impulse is 51.8 degree

part C)

as impulse = force/time

magnitude of average force = 57.7/0.0021

magnitude of average force = 27476 N

the magnitude of average force is 27476 N

part D)

direction of the average force is 51.8 degree

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