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Damm [24]
4 years ago
8

In an elastic collision, a 300 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

in the same direction. The initial speed of the leading bumper car is 5.30 m/s and that of the trailing car is 5.90 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds (in m/s)?
Physics
1 answer:
evablogger [386]4 years ago
3 0

Answer:

If we had:

v_{1i}=5.3m/s

v_{2i}=5.9m/s

We will have:

v_{1f}=5.9m/s

v_{2f}=5.3m/s

Explanation:

In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:

p_i=p_f

K_i=K_f

We will call our bumpers 1 and 2.

For the momentum equation we know that:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

Since all the masses are the same (300kg), they cancel out:

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

For the kinetic energy equation we know that:

\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}

Since all the masses are the same (300kg), they cancel out (and also the 2 dividing):

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

We then must solve this system:

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

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

Which we will rewrite as:

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

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

The last of these equations can be written as:

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

But we know that v_{1i}-v_{1f}=v_{2f}-v_{2i}, so those cancel out:

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

So we can write:

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

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

Which means:

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

Which solving for the final velocity leaves us with:

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

v_{1f}=v_{2i}

Grabbing any equation that relates both final velocities easily, for example v_{1i}-v_{1f}+v_{2i}=v_{2f}, we obtain:

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

So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):

v_{1f}=v_{2i}=5.9m/s

v_{2f}=v_{1i}=5.3m/s

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Answer:

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Explanation:

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The impulse-momentum theorem states the impulse on an object is equal to the change in momentum of that object. Momentum is given by p=mv. Since mass is constant, the train's change in momentum is:

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