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stellarik [79]
2 years ago
5

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc

ity of car B after the collision?
Physics
1 answer:
Nuetrik [128]2 years ago
4 0

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2

The data given are;

  • m1 = 281kg
  • u1 = 2.82m/s
  • m2 = 209kg
  • u2 = -1.72m/s
  • v1 = ?

Let's substitute the values into the equation.

v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
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Answer:

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Explanation:

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In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

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The initial potential energy of the spring is given by the equation:

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the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

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This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

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Does friction always oppose relative motion?
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Answer:

Yes

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3 0
3 years ago
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hjlf
The power that the light is able to utilize out of the supply is only 0.089 of the given.
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The energy required in this item is the product of the power utilized and the time. That is,
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