- 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.
it is generated by two objects pushing against each other with equal/almost equal force, therefore contracting against each other
Boundary separating two masses of air of different densities
The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
Learn more about potential energy:
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