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    Physics (Old) International Baccalaureate (IB) Practice Question: A football player kicks a stationary ball of mass 0.45 kg towards a wall. The initial speed of the ball after the kick i...

    A football player kicks a stationary ball of mass 0.45 kg towards a wall. The initial speed of the ball after the kick is 19 ms^-1 and the ball does not rotate. Air resistance is negligible and there is no wind.

    1. The player's foot is in contact with the ball for 55 ms. Calculate the average force that acts on the ball due to the football player.
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    Question
    SLPaper 2

    A football player kicks a stationary ball of mass 0.45 kg towards a wall. The initial speed of the ball after the kick is 19 ms^-1 and the ball does not rotate. Air resistance is negligible and there is no wind.

    1.

    The player's foot is in contact with the ball for 55 ms. Calculate the average force that acts on the ball due to the football player.

    [2]
    Verified
    Solution

    Δp=0.45×19\Delta p=0.45 \times 19Δp=0.45×19 OR a=190.055a=\frac{19}{0.055}a=0.05519​ ✔ «=F=0.45×190.055»160«N»✔«=F=\frac{0.45 \times 19}{0.055}» 160 «N» ✔«=F=0.0550.45×19​»160«N»✔

    2.

    The ball leaves the ground at an angle of 22°. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m. Calculate the time taken for the ball to reach the wall.

    [2]
    Verified
    Solution

    horizontal speed =19×cos⁡22«=17.6ms−1»✔=19 \times \cos 22 «=17.6 \mathrm{m} \mathrm{s}^{-1}» ✔=19×cos22«=17.6ms−1»✔ time =«distancespeed=1119cos⁡22=»0.62«s»✔= « \frac{\text{distance}}{\text{speed}}=\frac{11}{19 \cos 22}=» 0.62 «\mathrm{s}» ✔=«speeddistance​=19cos2211​=»0.62«s»✔

    3.

    The top of the wall is 2.4 m above the ground. Deduce whether the ball will hit the wall.

    [2]
    Verified
    Solution

    horizontal speed =19×cos⁡22«=17.6ms−1»✔=19 \times \cos 22 «=17.6 \mathrm{m} \mathrm{s}^{-1}» ✔=19×cos22«=17.6ms−1»✔ time =«distancespeed=1119cos⁡22=»0.62«s»✔= « \frac{\text{distance}}{\text{speed}}=\frac{11}{19 \cos 22}=» 0.62 «\mathrm{s}» ✔=«speeddistance​=19cos2211​=»0.62«s»✔

    4.

    In practice, air resistance affects the ball. Outline the effect that air resistance has on the vertical acceleration of the ball. Take the direction of the acceleration due to gravity to be positive.

    [2]
    Verified
    Solution

    vertical component of velocity =19sin⁡22«=7.1ms−1»✔=19 \sin 22 «=7.1 \mathrm{m} \mathrm{s}^{-1}» ✔=19sin22«=7.1ms−1»✔ s=ut+12at2s=ut + \frac{1}{2}at^2s=ut+21​at2 with u=7.1ms−1u=7.1 \mathrm{m} \mathrm{s}^{-1}u=7.1ms−1, a=−9.8ms−2a=-9.8 \mathrm{m} \mathrm{s}^{-2}a=−9.8ms−2 and t=0.62s✔t=0.62 \mathrm{s} ✔t=0.62s✔ s=«7.1×0.62−12×9.8×0.622=»2.2«m»✔s=«7.1 \times 0.62 - \frac{1}{2} \times 9.8 \times 0.62^2 =» 2.2 «\mathrm{m}» ✔s=«7.1×0.62−21​×9.8×0.622=»2.2«m»✔

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    5.

    The player kicks the ball again. It rolls along the ground without sliding with a horizontal velocity of 1.40 m s^-1. The radius of the ball is 0.11 m. Calculate the angular velocity of the ball. State an appropriate SI unit for your answer.

    [1]
    Verified
    Solution

    air resistance opposes «direction of» motion OR air resistance opposes velocity ✔ on the way up «vertical» acceleration is increased OR greater than g ✔ on the way down «vertical» acceleration is decreased OR smaller than g ✔

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