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sergiy2304 [10]

3 years ago

10

A 0.42 kg soccer ball is moving down field with a velocity of 12 m/s. A player kicks the ball so that it has a final velocity of

18 m/s down field. What is the change in the balls momentum? and find the constant force exerted by the players foot on the ball if the two are in contact for 0.020 s?

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

the change in momentum is 2.52 N·s

the constant force on the ball while in contact with the players foot is 126 N

Explanation:

p=momentum

m=mass

v=change in velocity

p=(0.42)(6)

p=2.52

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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

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Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

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At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

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$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

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We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

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$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

$g=9.8 m/s^{2}$ is the acceleratio due gravity

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Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

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$t=3.49 s$ (4)

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