Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)i^−(4.0m/s)j^. During the 3.00 msthat the racket and ball are in contact, the net force on the ball is constant and equal to −(380N)i^+(110N)j^.

A. What is the x -component of the impulse of the net force applied to the ball?
B. What is the y-component of the impulse of the net force applied to the ball?
C. What is the x-component of the final velocity of the ball?
D. What is the y-component of the final velocity of the ball?

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lcmendozaf lcmendozaf · Answer 1 · 2019-11-05T22:38:09+01:00

Answer:

A. Jx = -1.14N.s

B. Jy = 0.33N.s

C. Vfx = -0.36 m/s

D. Vfy = 1.89 m/s

Explanation:

The impulse is:

[tex]J = F_{neta}*\Delta t[/tex]

[tex]J = [-380,110]*0.003=[-1.14,0.33]Ns[/tex]

The components are:

Jx = -1.14N.s; Jy = 0.33N.s

The impulse is also calculated as:

J = m*ΔV = m*(Vf-Vo) where m = W/g = 0.056kg

Solving for Vf:

Vf = J/m + Vo = [-0.36, 1.89] m/s

The components are:

Vfx = -0.36 m/s; Vfy = 1.89 m/s

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