A player bounces a 043 mathrmkg soccer ball off head...

A player bounces a $0.43-\mathrm{kg}$ soccer ball off head, changing the velocity of the ball from $\overrightarrow{\mathbf{v}}_{1}=(8.8 \mathrm{m} / \mathrm{s}) \hat{\mathbf{x}}+(-2.3 \mathrm{m} / \mathrm{s}) \hat{\mathbf{y}}$ to $\overrightarrow{\mathbf{v}}_{\mathrm{f}}=(5.2 \mathrm{m} / \mathrm{s}) \hat{\mathbf{x}}+(3.7 \mathrm{m} / \mathrm{s}) \hat{\mathbf{y}} .$ If the ball is in contact with the player's head for 6.7 $\mathrm{ms}$ , what are (a) the direction and (b) the magnitude of the impulse delivered to the ball?

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Transcript

00:01 Question 22 asks us to find the magnitude and direction of the impulse delivered to a ball, but now we're moving in two dimensions.

00:07 So it says initially the ball is moving at 8 .8 meters per second in the x direction, that's what that tells us, right? this is a vector in unit vector notation, and negative 2 .3 meters per second in the y direction.

00:19 After contacting the player's head, the ball is moving at 5 .2 meters per second in x and 3 .7 meters per second in y.

00:27 And we're asked to find the impulse.

00:29 Remember, impulse is change in momentum, and this is equal to mass times change in velocity.

00:37 So, we can go ahead and find the change in velocity and multiply by mass in order to get this impulse.

00:44 The change in velocity, remember, is final minus initial.

00:48 But also remember, when you're dealing in multiple dimensions, you do need to keep your x's and y's separate.

00:54 So we can add x's, we can add y's, but we'll.

00:57 We can't add x as 2 ys unless we use the pythagorean theorem.

01:01 So our final x is 5 .2.

01:05 Our initial x is 8 .8.

01:08 So 5 .2 minus 8 .8 is negative 3 .6.

01:17 So the change in velocity in the x direction is negative 3 .6 meters per second.

01:25 Similarly in y, our final is 3 .7 meters per second...

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