A soccer player bounces a 0.430-kg soccer ball off of her head, which changes th

Question

A soccer player bounces a 0.430-kg soccer ball off of her head, which changes the velocity of the ball from: initial velocity vi = (8.8) x + (-2.3) y m/s to final velocity vf= (5.2) x + (3.7) y m/s ...where x and y represent unit vectors in the x and y directions, respectively. During the bounce, the ball was in contact with the player's head for 6.7 ms.

What was the magnitude of the impulse delivered to the ball?

What is the direction of the impulse delivered to the ball? Specify the angle in degrees, as a positive number between 0 and 180º, measured counterclockwise from the positive x axis.

What was the magnitude of the average force exerted by the ball on her head?

Explanation / Answer

Impulse = change in momentum

= m ( vf - vi )

= 0.430 [ ( 5.2x + 3.7y ) - (8.8x -2.3y ) ] = 0.430(-3.6x + 6y )

Impulse = - 1.548 x + 2.58y

magnitude = sqrt(1.548^2 + 2.58^2) = 3 kg m/s .........Ans

angle =180 - tan^-1 (2.58 / 1.548 ) = 120.96 deg .........Ans

Impulse = Force x time

3 = F x (6.7 x 10^-3 )

F= 447.76 N

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