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Vaselesa [24]
3 years ago
7

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
Physics
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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