A soccer ball player bounces the ball off her head, changing the velocity of the ball. She changes the x compo?
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 = 9.9 m/s to vfx = 3.6 m/s and the y component from viy = −1.0 m/s to vfy = 4.6 m/s. If the ball has a mass of 0.41 kg and is in contact with the player's head for 7.1 ms, determine the following. (a) direction of the impulse delivered to the ball ° counterclockwise from the +x-axis (b) magnitude of the impulse delivered to the ball kg · m/s
Physics Help! A soccer ball player bounces the ball off her head, changing the velocity of the ball.?
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 = 8.2 m/s to vfx = 3.9 m/s and the y component from viy = −2.5 m/s to vfy = 4.4 m/s. If the ball has a mass of 0.41 kg and is in contact with the player's head for 7.1 ms, determine the following. (a) direction of the impulse delivered to the ball _____ ° counterclockwise from the +x-axis (b) magnitude of the impulse delivered to the ball ____ kg · m/s
Physics. PLEASE HELP ASAP! A soccer player bounces the ball off her head, changing the velocity of the ball.?
Impulse = Force * time AND Impulse = Change of momentum Since we know the mass and change of velocity, we can determine of momentum in both directions. x = 0.430 * (5.5 – 8.5) = -1.29 y = 0.430 * (4.4 – -3.3) = 3.311 ∆M = √(-1.29^2 + 3.331^2) = √12.62821 The change of the ball’s momentum is approximately 3.55 kg * m/s. Let’s use this to determine the magnitude of force. F * 0.0071 = √12.62821 F = √12.62821 ÷ 0.0071 The force is approximately 500.5 N. To determine the angle above the negative x axis, use the following equation. Tan θ = y ÷ x = 3.311 ÷ -1.29 This is approximately 68.7˚ above the negative x axis. I hope this is helpful for you.
A soccer ball player bounces the ball off her head, changing the velocity of the ball. She changes the x compo?
Change in horizontal (x-axis) velocity .. ∆vx = (vfx - vix) = 4.80 - 8.10 .. .. ∆vx = - 3.30 m/s Change in vertical (y) velocity .. ∆vy = (vfy - viy) = 5.50 - (-2.90) .. .. ∆vy = 8.40 m/s Using Pythagoras to find ball's resultant velocity change ∆v .. (∆v)² = 3.30² + 8.40² = 81.45 .. .. ∆v = 9.0 m/s (b) Impulse on ball = change in mom / time = m.∆v / t = 0.43kg x 9.0m/s / 7.10^-3s .. .. .. ►Impulse = 545.10 kg.m/s (a) Impulse direction = direction of resultant velocity change ∆v ∆vx = - 3.30m/s (←) ∆vy = + 8.40 m/s (↑) Clockwise angle from horizontal .. tanθ (dy/dx) = 8.40 / 3.30 = 68.6° Anti-clockwise angle = 180° - 68.6° .. .. ►Angle = 111.40°
A player bounces a 0.43 kg soccer ball off her head?
Calculate the initial momentum (Pi): Pi = √((0.43x8.8)^2 + (0.43x2.3)^2) = Pi = √((3.8)^2 + (0.99)^2) = √ (14.3 + 0.98) = 3.9 Ns (284.6º bearings) Calculate the final momentum (Pf): Pf = √(0.43x5.2)^2 + (0.43x3.7)^2) = √(2.24)^2 + (1.59)^2) = √ (5.02 + 2.53) = 2.75 Ns ( 35.4º bearings) Impulse = Ft = ∆P Where F = average force t = time of contact ∆P = change of momentum Now ∆P = Pf – Pi = Pf + (-Pi) But you need to do a vector addition. You’ll need the cosine rule to determine the resultant vector: ( ∆P ^2 ) = Pf^2 + Pi^2 – 2 x Pf x Pi x cos (14.6 + 35.4) ( ∆P ^2 ) = 2.75^2 + 3.9^2 – 2 x 2.75 x 3.9 x cos (50) ( ∆P ^2 ) = 7.56 + 15.28 – 21.45 x 0.64 ( ∆P ^2 ) = 22.84 – 13.79 ( ∆P ^2 ) = 9.05 ∆P = 3.0 Ns Answers: a) Answer = ∆P = impulse = 3.0 Ns b) Direction = (bearing 339.2º) (Use sine rule & geometry ) Average force (F) can be now calculated: Ft = ∆P F = ∆P / t = 3.0 / 6.7E-3 = 449 Newtons NB: check my working.