# Impulse of deflected ball

1. Feb 23, 2008

### physicsgrouch

[SOLVED] Impulse of deflected ball

1. Recent studies have raised concern about heading' in youth soccer (i.e., hitting the ball with the head). A soccer player heads' a 0.421 kg ball, deflecting it by 50.0 degrees, and keeps its speed of 10.40m/s constant. (The deflection angle is the angle between the ball's initial and final velocity vectors.) What is the magnitude of the impulse which the player must impart to the ball?

m = 0.421 kg
v$$_{}i$$ = -10.4 m/s
v$$_{}f$$ = 10.4 m/s at 50 degrees

2. The equations

impulse = $$\Delta$$p = m(v$$_{}i$$) - m(v$$_{}f$$)

so impulse = m(v$$_{}f$$ - v$$_{}i$$)

So I multiplied the mass by the change in velocity. Namely:

3. The solution

0.421 *{sqrt[ (10.40*sin50)^2 + (10.40*(cos50 +1))^2) ]}

So I got about 7.936. But this is wrong. What's up?

2. Feb 23, 2008

### foxjwill

I think you typed it into the calculator wrong. I just plugged it into google and got
Code (Text):

.421 * sqrt(((10.40 * sin(50))^2) + ((10.40 * (cos(50) + 1))^2)) = 8.67976478

or, in $$\LaTeX$$

$$.421 \sqrt{ \left(10.4\sin{50}\right)^2 + \left(10.4\left(\cos{50}+1\right)\right)^2 } = 8.67976478$$

3. Feb 23, 2008

### physicsgrouch

I think that might be in radians...

Thanks, though. My teacher went over the homework, and it seems that I was measuring the wrong angle. The angle is SUPPOSED to be between the tails of the two vectors, but I was measuring the angle between the tip of the initial and the tail of the final. So my calculations should have treated the angle as 130 degrees. Alternatively, I could have made both vectors' x-components have the same signs.

Thanks anyway.