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3 years ago

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A 0.180-kilogram car traveling at 0.80 m/s to the right collides with a 0.100-kilogram cart intially at rest. The carts lock tog

ether upon collison. Calculate the final velocity of the carts

1 answer:

prohojiy [21]3 years ago

6 0

0.51 m/s to the right.

<h3>Explanation</h3>

The two carts lock together. As a result, the collision is inelastic. Kinetic energy will not conserve. Still, momentum conserves.

What's the momentum <em>p</em> of the two carts?

Before the collisions:

$p_1 = m_1 \cdot v_1 = 0.180 \times 0.80 = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}$ ;
$p_2 = m_2 \cdot v_2 = 0$ .

Sum of momentum: $p = p_1 + p_2 = 0.144\; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}$ .

Momentum conserves. As a result, $p(\text{after collision}) = p(\text{before collision}) = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}$

Velocity is the same for the two carts after the collision. Let $v$ denote that velocity.

$p(\text{after collision}) = m_1\cdot v+ m_2 \cdot v = (m_1 + m_2) \cdot v$ .

$v = \dfrac{p(\text{after collision})}{m_1 + m_2} \\ \phantom{v} = \dfrac{0.144}{0.180 + 0.100} \\ \phantom{v} = 0.51 \; \text{m}\cdot \text{s}^{-1}$ .

Direction of the movement will stay the same. Both cars are now moving to the right.

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By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

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$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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