Just before it is struck by a racket, a tennis ball weighing0.560 N has a veloci

Question

Just before it is struck by a racket, a tennis ball weighing0.560 N has a velocity of (20.0m/s)i^(4.0m/s)j^. During the 3.00 ms that 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?

Explanation / Answer

Impulse on the ball = force * time = [-(380)i+(110)j]*3*e-3 Ns. = [-(1.14)i+(0.33)j] Ns

(a) x-component of the impulse = -1.14 Ns

(b) y-component of the impulse = 0.33 Ns

Now, final velocity can be determined by applying the formula impulse = change in linear momentum

i.e. F*t = m (vf - vi)

here, m = 0.560/9.81 kg = 0.057 kg

So, [-(1.14)i+(0.33)j] = 0.057*[vf - (20.0)i+(4.0)j]

=> vf - (20.0)i+(4.0)j = -20i + 5.79j

=> vf = 1.79j

(c) x-component of the final velocity of the ball = 0

(d) y-component of the final velocity of the ball = 1.79 m/s.

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