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Usimov [2.4K]
3 years ago
11

A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)
Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

v_2 = 23.182m/s

Explanation:

Given

m_1 = 0.17kg

m_2 = 0.11kg

u_1 = 15m/s -- Initial Velocity of the first

u_2 = 0m/s -- Initial Velocity of the second

v_1 = 0m/s -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck (v_2)

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 ---- law of conservation of momentum

Substitute in the right values

0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2

0.17 * 15 + 0 = 0 + 0.11*v_2

2.55 = 0.11*v_2

Solve for v2

v_2 = 2.55/0.11

v_2 = 23.182m/s

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

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also by division of above two equations we have

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3 0
3 years ago
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Answer:

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8 0
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hjlf

Answer:

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Explanation:

The kinetic energy of the block is equal to potential energy of spring at maximum compression

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m is mass of block , V is its velocity , K is spring constant and  X is maximum compression or its amplitude.

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