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topjm [15]
2 years ago
15

A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.

0 m/s after the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface, what is the final velocity of the canoe?
Physics
1 answer:
juin [17]2 years ago
6 0
  • A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.0 m/s.
  • After the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface.
  • Mass of the canoe (m1) = 16 Kg
  • Mass of the raft (m2) = 14 Kg
  • Initial velocity of the canoe (u1) = 12.5 m/s
  • Initial velocity of the raft (u1) = - 16 m/s [Here, the raft's velocity is negative, because the objects are moving in the opposite direction]
  • Total momentum of the system = m1u1 + m2u2 = [(16 × 12.5) + (14 × -16)] Kg m/s = (200 - 224) Kg m/s = -24 Kg m/s
  • Final velocity of the raft (v2) = 14.4 m/s
  • Let the final velocity of the canoe be v1.
  • Total momentum of the system after the impact = m1v1 + m2v2 = [(16 × v1) + (14 × 14.4)] Kg m/s = 16v1 Kg + 201.6 Kg m/s
  • According to the law of conservation of momentum, Total momentum of the system before the impact = Total momentum of the system after the impact
  • or, -24 Kg m/s = 16v1 Kg + 201.6 Kg m/s
  • or, -24 Kg m/s - 201.6 Kg m/s = 16v1 Kg
  • or, -225.6 Kg m/s = 16v1 Kg
  • or, v1 = -225.6 Kg m/s ÷ 16 Kg
  • or, v1 = -14.1 m/s

<u>Answer:</u>

<u>T</u><u>he final velocity of the </u><u>canoe </u><u>is </u><u>-</u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>or </u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>to </u><u>the </u><u>right.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to t
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Explanation:

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radius of wheel, R = 0.65 m

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Moment of inertia = Moment of inertia of disc + moment f inertia of the clay

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          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

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