Figure caption: A puck sliding to the right with a velocity of 4 meters per second. The puck approaches an incline that rises up 1 meter before leveling off again.

A hockey puck sliding on smooth ice at 4.5 m/s comes to a 1-m high hill. How fast is the puck going at the top of the hill?

Question

Figure caption: A puck sliding to the right with a velocity of 4 meters per second. The puck approaches an incline that rises up 1 meter before leveling off again.

A hockey puck sliding on smooth ice at 4.5 m/s comes to a 1-m high hill. How fast is the puck going at the top of the hill?

BlackTom AI · Accepted Answer

1. Conservation of mechanical energy applies: on smooth (frictionless) ice, mechanical energy (kinetic + gravitational potential) is conserved. Key concepts: KE = 1/2 m v^2, PE = m g h, g ≈ 9.8 m/s^2. Because both KE and PE are proportional to mass m, the mass cancels and is not needed. The angle of the hill does not matter for energy balance.

2. Reasoning (step by step): set initial total energy equal to energy at the top of the 1-m hill. Let v0 = 4.5 m/s (initial speed), v = speed at top, h = 1.0 m.

3. Calculations with intermediate steps:
- Initial kinetic energy: KE0 = 1/2 m v0^2 = 1/2 m (4.5^2) = 1/2 m (20.25) = 10.125 m (Joules per kg factor m).
- Initial potential energy: PE0 = m g (0) = 0 (take base as zero).
- Energy at top: KE_top = 1/2 m v^2, PE_top = m g h = m (9.8)(1.0) = 9.8 m.
- Conservation: KE0 + PE0 = KE_top + PE_top
  => 1/2 m v0^2 = 1/2 m v^2 + m g h
  Divide both sides by m: (1/2) v0^2 = (1/2) v^2 + g h
  Multiply by 2: v0^2 = v^2 + 2 g h
  Solve for v^2: v^2 = v0^2 - 2 g h
- Plug numbers: v^2 = (4.5^2) - 2(9.8)(1.0) = 20.25 - 19.6 = 0.65
- v = sqrt(0.65) = 0.80623... m/s ≈ 0.81 m/s

4. Why this is correct: energy lost as gravitational potential (m g h = 9.8 m per kg) comes from initial kinetic energy. Because the remaining kinetic energy corresponds to v^2 = 0.65, the speed must be sqrt(0.65) ≈ 0.81 m/s.

5. Why other choices are incorrect:
- "The puck does not make it to the top of the hill.": false — v^2 is positive (0.65), so it reaches the top with speed ≈0.81 m/s.
- "4.4 m/s": far too large; would imply almost no loss of speed and ignores the 1-m rise.
- "Can't answer without knowing the mass of the puck.": false — mass cancels in the energy equation.
- "Can't say without knowing the angle of the hill.": false — energy methods do not require the slope angle.

6. Conclusion: The calculated speed at the top is about 0.81 m/s, which matches the provided correct answer 0.81 m/s.

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