What is the direction of impulse delivered to the ball?

[ajmCane22]

Homework Statement

A player bounces a 0.47 kg soccer ball off her head, changing the velocity of the ball from vi= (8.8 m/s)x + (-2.4 m/s)y
to vf= (5.0 m/s)x + (3.0 m/s)y.

If the ball is in contact with the player's head for 6.7 ms, what is the direction of the impulse delivered to the ball?

The Attempt at a Solution

I tried finding the change in velocity which I thought would be (vf-vi)x + (vf-vi)y = (5.0-8.8)x + (3+2.4) = (-3.8 m/s)x + (5.4 m/s)y, so theta would be tan^-1 of 5.4/-3.8 which equals -54.9 degrees. That answer is wrong (I also divided by .0067s which gives the same result, and I also tried 360-54.9=305.1). What am I doing wrong? Please help. Thanks in advance.

 

[Doc Al]

I tried finding the change in velocity which I thought would be (vf-vi)x + (vf-vi)y = (5.0-8.8)x + (3+2.4) = (-3.8 m/s)x + (5.4 m/s)y, so theta would be tan^-1 of 5.4/-3.8 which equals -54.9 degrees.

Draw yourself a picture of that change in velocity vector. Then, after doing your trig, express the angle with respect to the positive x axis.

 

[ajmCane22]

I drew a picture and the vector goes through the x-axis. I'm not sure how that helps me. Is the trig correct, because maybe I just don't get what is meant by "with respect to the positive x-axis."

 

[gneill]

Your ΔV vector has components (-3.8, 5.4). On your diagram that vector should extend from the origin to the point (-3.8, 5.4). If you were standing at the origin and pointing your finger out along the direction of the X-axis, then swung your arm up and over to point along the direction of the ΔV vector, through what angle would your arm sweep?

 

[Doc Al]

I drew a picture and the vector goes through the x-axis. I'm not sure how that helps me. Is the trig correct, because maybe I just don't get what is meant by "with respect to the positive x-axis."

An answer like -54.9 degrees doesn't mean much unless you specify with respect to what. What quadrant does your vector lie in? (Show your diagram.)

(Don't just plug into a formula blindly. First draw a diagram, then use a bit of trig.)

 

[ajmCane22]

omg, I get it. Thank you both so much. One more thing, so then when trying to find the impulse, shouldn't it be pi-pf = mvf-(-mvi) = 0.47*5.83-(-0.47*9.12) = 7.02
that doesn't seem to be right. I don't understand what I'm doing wrong.

 

[gneill]

Impulse is a vector quantity. You'll have to deal with the components. You can determine the magnitude of the impulse vector when the smoke clears.

 

[Doc Al]

One more thing, so then when trying to find the impulse, shouldn't it be pi-pf = mvf-(-mvi) = 0.47*5.83-(-0.47*9.12) = 7.02
that doesn't seem to be right. I don't understand what I'm doing wrong.

Use the Δv you already calculated using components. What's the magnitude of that vector?

 

[ajmCane22]

I calculated 6.603, so would the impulse be 6.603*mass = 6.603*0.47 = 3.10?

 

[Doc Al]

I calculated 6.603, so would the impulse be 6.603*mass = 6.603*0.47 = 3.10?

Sounds good to me.

The important point is that since momentum is a vector, you cannot just subtract the magnitudes of the initial and final momenta to get the change. You must take their directions into account. (Of course, using components automatically takes account of direction.)