Physics Question- HELP!?
A soccer ball player bounces the ball off her head, changing the velocity of the ball. She changes the x-component of the velocity of the ball from vix = 7.9 m/s to vfx = 5.9 m/s and the y-component from viy =−1.4 m/s to vfy = 4.2 m/s. If the ball has a mass of 0.45 kg and is in contact with the player's head for 7.1 ms, determine the following.
(a)direction of the impulse (in degrees counterclockwise from the +x-axis) delivered to the ball° counterclockwise from the +x-axis
magnitude of the impulse (in kg · m/s) delivered to the ball kg · m/s
- Old Science GuyLv 79 months ago
impulse = change in momentum
delta(p) = m delta(v) = (0.45(-2.0) + 0.45(5.6) i)
I'll use a vector calculator to convert this to R, theta
(-0.90 + 25.2 i) = 25.2 @ 92°
magnitude is 25.2 kg*m/s
and direction is 92° CCW from +x-axis or 2° past vertical
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- nyphdinmdLv 79 months ago
Pi = m*(vix i_hat + viy j_hat)
Pf = m*(vfx i_hat + vfy j_hat)
DP = Pf - Pi = m*((vfx - vix) i_hat + (vfy - viy) j_hat)
Impulse I = DP/dt = 63.38 kg/s ((vfx - vix) i_hat + (vfy - viy) j_hat)
= 63.38 kg/s*(-2 m/s i_hat + 5.6 m/s j_hat) = -126.8 i_hat + 354.9 j_hat N
Angle wrt to x - axis of I: q = arctan(354.9/(-126.8) = -70.35 degrees
magnitude of I = sqrt((-126.8)^2 +354.9^2) = 376.9 N