Coach Johnson\'s bat exerts a horizontal force on a 0.145 kg baseball of F v e c

Question

Coach Johnson's bat exerts a horizontal force on a 0.145 kg baseball of

Fvec=[(1.60×107N/s)t(6.00×109N/s2)t2]i^

between t=0 and t=2.50ms. At t=0 the baseball's velocity is (40.0i^+5.0j^)m/s. For this problem, consider the positive x direction to be directed away from Coach Johnson and the positive y direction to be directed upward.

Part A

Calculate the impulse exerted by the bat on the ball during the 2.50 msthat they are in contact.

Express your answer to four significant figures. Use i^ and j^ for the unit vectors in the x and ydirections respectively.

ANS: 18.75i^ kg*m/s

Part B

Calculate the impulse exerted by gravity on the ball during this time interval.

Express your answer to three significant figures. Use i^ and j^ for the unit vectors in the x and ydirections respectively.

ANS: -0.0036j^ kg*m/s

Part C

Calculate the average force exerted by the bat on the ball during this time interval.

Express your answer to three significant figures. Use i^ and j^ for the unit vectors in the x and ydirections respectively.

Part D

Calculate the momentum of the baseball at t =2.50ms.

Express your answer to three significant figures. Use i^ and j^ for the unit vectors in the x and ydirections respectively.

Part E

Calculate the velocity of the baseball at t =2.50ms.

Express your answer to three significant figures. Use i^ and j^ for the unit vectors in the x and ydirections respectively.

So I managed to get A and B... but I keep getting C through E incorrect...

Explanation / Answer

A)

impulse = J = integration(F*dt)

J = [(1.60×10^7)t^2/2(6.00×10^9)t^3/3]

J = [(1.60×10^7)t^2/2(6.00×10^9)t^3/3]

J = (1.6*10^7*(2.5*10^-3)^2/2 - (6*10^9*(2.5*10^-3)^3/3) i

J = 18.75i

B)

J = Fg*t = -0.145*9.81*2.5*10^-3 = -0.0036 kg m/s

(c)

Fav = Fo + F2.5ms /2

Fav = ((1.6*10^7*(2.5*10^-3) - (6*10^9*(2.5*10^-3)^2))/2 i

Fv = 1250 N

D)

J = Pf - Pi

J = Pf -Pi

Pf = J + Pi

Pf = 18.75i -0.145*(40i+5j)

Pf = 18.75i - 5.8i - 0.725 j

Pf = 12.95i - 0.725 j

E)

Pf = m*Vf = 89.3i - 5j

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