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Just before it is struck by a racket, a tennis ball weighing 0.560 $\mathrm{N}$ has a velocity of $(20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\mathrm{J}}$ . During the

3.00 $\mathrm{ms}$ that the racket and ball are in contact, the net force on the ball is constant and equal to $-(380 \mathrm{N}) \hat{\imath}+(110 \mathrm{N}) \mathrm{J}$ . (a) What are the $x$ -and $y$ components of the impulse of the net force applied to the ball? $(b)$ What are the $x-$ and $y$ -components of the final velocity of the ball?

(a) $J_{x}=-1.14 \mathrm{N.S} \quad-\quad J_{y}=0.33 \mathrm{N.s}$

(b) $v_{2 x}=0.05 \mathrm{m} / \mathrm{s} \quad-\quad v_{2 y}=1.775 \mathrm{m} / \mathrm{s}$

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{'transcript': "problem. 8.65. So we have a tennis ball with a given wait. Careful to note, this is the weight, not the mass. It has an initial velocity of this right before it gets hit by a tennis racket, which it's in contact for with three milliseconds, and it exerts a constant force that's given here. No, we'd like to do is find out what is the impulse that this force produces. And then what is the new velocity of the tennis ball After this, forces acted the part. So before we get anywhere else, let's just real quick. Be careful and divide the weight were given by the acceleration due to gravity. And we find this has a mass of 0.57 one kilograms now for part A. We don't need that. But for part A, we have the, uh, The impulse for a constant or average force is equal just to that force times the amount of time to spend. So we take our force here, multiply it by three times 10 to the negative third seconds, and we end up with negative one point on four Newton seconds and the extraction and zero it's a parenthesis 0.3 30 Newton seconds in what direction? So now we need to use the fact that the impulse is also equal to the change in momentum. So the final Momenta Mme. Is equal to the initial momentum, plus the change. So now we have m. It's the final. The thing we'd like to know equals m times be initial forgiven up here, plus our horse times the length of time it acts. So you divide through by m but in the numbers and we find that this is 0.0 four meters per second and the extraction plus 1.8 meters per second in the white direction."}