Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 16P

(a)

To determine

The change in momentum (magnitude and direction) of the baseball.

(a)

Expert Solution
Check Mark

Answer to Problem 16P

The magnitude of momentum of baseball is 3.8kgm/s and it directed 37° above the horizontal direction opposite vi.

Explanation of Solution

Take direction opposite to initial velocity of baseball as +x direction and upward direction as +y direction.

Write the expression for the magnitude of change in momentum along x direction.

Δpx=m(vfxvix)

Here, Δpx is the change in momentum along x direction, m is the mass of baseball, vfx is the final velocity of baseball along x direction and vix is the initial velocity of baseball along x direction.

Write the expression for the magnitude of change in momentum along y direction.

Δpy=m(vfyviy)

Here, Δpy is the change in momentum along y direction, m is the mass of baseball, vfy is the final velocity of baseball along y direction and viy is the initial velocity of baseball along y direction.

Write the expression for the magnitude of total change in momentum.

|Δp|=(Δpx)2+(Δpy)2

Here, |Δp| is the magnitude of change in momentum of total momentum, Δpx is the magnitude of change in momentum along x direction and Δpy is the magnitude of change in momentum along y direction.

Substitute m(vfxvix) for Δpx and m(vfyviy) for Δpy in above equation to get |Δp|.

|Δp|=m(vfxvix)2+(vfyviy)2 (I)

Write the expression for direction of momentum.

θ=tan1(m(vfyviy)m(vfxvix))=tan1((vfyviy)(vfxvix)) (II)

Here, θ is the angel that momentum vector makes with x axis.

Conclusion:

Substitute 0.15kg for m , 0m/s for vfx, 20m/s for vix , 15m/s for vfy and 0m/s for vfy in equation (I) to get

|Δp|=0.15kg((0m/s)(20m/s))2+(15m/s0m/s)2=3.8kgm/s

Substitute 0.15kg for m , 0m/s for vfx, 20m/s for vix , 15m/s for vfy and 0m/s for vfy in equation (II) to get θ.

θ=tan1((15m/s0m/s)(0m/s)(20m/s))=37°

Therefore, the magnitude of momentum of baseball is 3.8kgm/s and it directed 37° above the horizontal direction opposite vi.

(b)

To determine

The average force of the bat on the ball.

(b)

Expert Solution
Check Mark

Answer to Problem 16P

The average force of the bat on the ball is 75N in the same direction as Δp.

Explanation of Solution

Write the expression for the average force of the bat on the ball.

Fav=ΔpΔt

Here, Fav is the average force of the bat on the ball, Δp is the momentum change

Conclusion:

Substitute 3.75kgm/s for Δp and 50ms for Δt in above equation to get Fav.

Fav=3.75kgm/s50ms×103s1ms=75N

Therefore, the average force of the bat on the ball is 75N in the same direction as Δp.

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Chapter 7 Solutions

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