Bumper car $$ 1(281 \mathrm{~kg}) $$ moves east at $$ 2.82 \mathrm{~m} / \mathrm{s} $$. Bumper car $$ 2(209 \mathrm{~kg}) $$ moves west at $$ 1.72 \mathrm{~m} / \mathrm{s} $$. If the cars have an elastic collision, what would the velocity of car 2 be afterwards? $$ \mathrm{v}_{2 \mathrm{f}}=[?] \mathrm{m} / \mathrm{s} $$ Remember: East is positive $$ (+) $$; west is negative $$ (-) $$.

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We use the formula for a one-dimensional elastic collision. For two objects we have

$$
v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}+\frac{m_2-m_1}{m_1+m_2}v_{2i}
$$

where

$$
m_1=281\text{ kg},\quad v_{1i}=+2.82\text{ m/s},
$$
$$
m_2=209\text{ kg},\quad v_{2i}=-1.72\text{ m/s}
$$

First, compute the total mass:

$$
m_1+m_2=281+209=490\text{ kg}
$$

Now plug in the values:

$$
v_{2f}=\frac{2(281)}{490}(2.82)+\frac{209-281}{490}(-1.72)
$$

Calculate the factors step by step.

1. For the first term:

$$
\frac{2(281)}{490}=\frac{562}{490}\approx1.14694
$$

So

$$
1.14694\times2.82\approx3.23488\text{ m/s}
$$

2. For the second term:

$$
209-281=-72,\quad\frac{-72}{490}\approx-0.14694
$$

Thus

$$
-0.14694\times(-1.72)\approx0.25270\text{ m/s}
$$

Now sum the two terms:

$$
v_{2f}\approx3.23488+0.25270\approx3.48758\text{ m/s}
$$

Since the result is positive, car 2 moves east after the collision.

Thus, the final velocity of car 2 is

$$
v_{2f}\approx3.49\text{ m/s}
$$
After the elastic collision, car 2 moves east at approximately $$ 3.49 \, \text{m/s} $$.

Bumper car moves east at .
Bumper car moves west at .
If the cars have an elastic collision,
what would the velocity of car 2 be
afterwards?
Remember: East is positive ; west is negative .

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Bumper car moves east at . Bumper car moves west at . If the cars have an elastic collision, what would the velocity of car 2 be afterwards? Remember: East is positive ; west is negative .