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Daniel [21]
2 years ago
10

Two billiard balls with the same mass m move towards one another. Ball one travels in the positive x-direction with a speed of v

1i, and ball two travels in the negative x-direction with a speed of v2i. The two balls collide elastically, and both balls change direction after the collision. If the initial speeds of the balls were v1i=2.0m/s and v2i =1.0m/s, what would be the final speed and direction of ball two, v2f, in m/s?
Physics
1 answer:
Kitty [74]2 years ago
4 0

Answer:

v_{2f}=2m/s in the positive x-direction.

Explanation:

Conservation of momentum tells us that the momentum before the collision must be equal to the momentum after the collision: p_i=p_f.

Before the collision we have:

p_i=m_1v_{1i}+m_2v_{2i}=mv_{1i}+mv_{2i}=m(v_{1i}+v_{2i})

And after the collision:

p_f=m_1v_{1f}+m_2v_{2f}=mv_{1f}+mv_{2f}=m(v_{1f}+v_{2f})

Since they must be equal we have:

m(v_{1i}+v_{2i})=m(v_{1f}+v_{2f})

Which is the same as:

v_{1i}+v_{2i}=v_{1f}+v_{2f}

And can be written as:

v_{1i}-v_{1f}=v_{2f}-v_{2i}

Since the collision is elastic, kinetic energy is conserved: K_i=K_f

Before the collision we have:

K_i=\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{mv_{1i}^2}{2}+\frac{mv_{2i}^2}{2}=\frac{m}{2}(v_{1i}^2+v_{2i}^2)

And after the collision:

K_f=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}=\frac{mv_{1f}^2}{2}+\frac{mv_{2f}^2}{2}=\frac{m}{2}(v_{1f}^2+v_{2f}^2)

Since they must be equal we have:

\frac{m}{2}(v_{1i}^2+v_{2i}^2)=\frac{m}{2}(v_{1f}^2+v_{2f}^2)

Which is the same as:

v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2

And can be written as:

v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2

Which is the same as:

(v_{1i}-v_{1f})(v_{1i}+v_{1f})=(v_{2f}-v_{2i})(v_{2f}+v_{2i})

But since we already proved that v_{1i}-v_{1f}=v_{2f}-v_{2i}, we can cancel those terms and we have that:

v_{1i}+v_{1f}=v_{2f}+v_{2i}

So we have the system:

v_{1i}-v_{1f}=v_{2f}-v_{2i}

v_{1i}+v_{1f}=v_{2f}+v_{2i}

Adding both sides between them we obtain:

v_{1i}-v_{1f}+v_{1i}+v_{1f}=v_{2f}-v_{2i}+v_{2f}+v_{2i}

Which is:

2v_{1i}=2v_{2f}

Which means:

v_{2f}=v_{1i}=2m/s

Where we used a positive velocity since ball 1 travels in the positive x-direction at the beginning.

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c) 53.13°

d) 106.67 m

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c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

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cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

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