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A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling hor

ID: 1349617 • Letter: A

Question

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 55.0 m/s , and it leaves the bat traveling to the left at an angle of 30 above horizontal with a speed of 60.0 m/s . The ball and bat are in contact for 1.85 ms .

Part A

Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right

Express your answer using two significant figures.

8383

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Part B

Find the vertical component of the average force on the ball.

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 55.0 m/s , and it leaves the bat traveling to the left at an angle of 30 above horizontal with a speed of 60.0 m/s . The ball and bat are in contact for 1.85 ms .

Part A

Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right

Express your answer using two significant figures.

Fx =

8383

  N  

SubmitMy AnswersGive Up

Incorrect; Try Again; 4 attempts remaining

Check your signs.

Part B

Find the vertical component of the average force on the ball.

Explanation / Answer


mass = m = 0.145 kg


v1x = +55 m/s


v2x = -v2*cos30 = -60*cos30 = -52 m/s


change in momentum along x

dPx = P2x - P1x = m*v2x - m*v1x = m*(v2x-v1x)

Fx = dPx/dt = m*(v2x-v1x)/dt = 0.145*(-52-55)/(1.85*10^-3)


Fx = -8386.5 N


part(B)

along vertical


V1y = 0


v2y = v2*sin30 = 60*sin30 = 30


dPy = P2y-P1

Fy = dPy/dt = m*(v2y-v1y)/t = 0.145*(30-0)/(1.85*10^-3)


Fy = 2351.3 N

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