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Midterm II Practice

This collection features solved problems covering work and energy, linear momentum and its conservation, and rotational dynamics—including rotational motion, moments of inertia, and angular momentum.

Problem 3 — The Tangled Ice Gliders

2D momentum conservation in an inelastic collision, followed by a 1D elastic collision — tracking vectors and energy through both.

Two hockey-style pucks slide on a flat, frictionless horizontal sheet of ice.

  • Puck 1 (\(m_1 = 5\text{ kg}\)) slides due East at a constant speed of \(8\text{ m/s}\).
  • Puck 2 (\(m_2 = 5\text{ kg}\)) slides at \(10\text{ m/s}\) at an angle of \(37^\circ\) North of West.

The two pucks collide at the origin (Use \(\sin 37^\circ = 0.6\) and \(\cos 37^\circ = 0.8\)).

The Completely Inelastic Collision. During the collision, Velcro on the pucks' edges causes them to stick together, forming a single combined object of mass \(M\).

  1. Resolve the initial velocity of Puck 2 into its \(x\) (East–West) and \(y\) (North–South) components.
  2. Calculate the final velocity vector (magnitude and direction) of the combined pucks immediately after the collision.
  3. Calculate the total kinetic energy lost during this collision.
  4. The Elastic Collision. The combined \(M = 10\text{ kg}\) mass from Part A (moving with its final velocity from Part A) slides along the ice and undergoes a head-on, perfectly elastic 1D collision with a third stationary puck, Puck 3 (\(m_3 = 5\text{ kg}\)). Calculate the final velocities of the combined mass (\(M = 10\text{ kg}\)) and Puck 3 (\(m_3 = 5\text{ kg}\)) after this elastic collision.
  5. Prove mathematically that kinetic energy was conserved during this second collision.
I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

- Topic: Momentum conservation, and inelastic and elastic collisions.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Vector projections, and energy conservation during collisions.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Convention. East \(= +x\), North \(= +y\).

Part 1 — Resolve Puck 2's velocity into components

Puck 2 moves at \(10\text{ m/s}\), \(37^\circ\) North of West. West is \(-x\), North is \(+y\):

\[v_{2x} = -10\cos 37^\circ = -10(0.8) = -8\text{ m/s}\]
\[v_{2y} = +10\sin 37^\circ = +10(0.6) = +6\text{ m/s}\]

Part 2 — Final velocity after sticking (\(M = 10\text{ kg}\))

Conserve momentum in each direction:

\[p_x = m_1 v_{1x} + m_2 v_{2x} = 5(8) + 5(-8) = 0\]
\[p_y = m_1 v_{1y} + m_2 v_{2y} = 5(0) + 5(6) = 30\ \text{kg·m/s}\]

The \(x\)-momenta cancel exactly. Final velocity:

\[V_x = \frac{p_x}{M} = 0, \qquad V_y = \frac{p_y}{M} = \frac{30}{10} = 3\text{ m/s}\]
\[\boxed{V = 3\text{ m/s, directed due North}}\]

Part 3 — Kinetic energy lost

Initial:

\[KE_i = \tfrac{1}{2}(5)(8)^2 + \tfrac{1}{2}(5)(10)^2 = 160 + 250 = 410\text{ J}\]

Final:

\[KE_f = \tfrac{1}{2}(10)(3)^2 = 45\text{ J}\]
\[\Delta KE = 410 - 45 = \boxed{365\text{ J lost}}\]

Part 4 — Elastic 1D collision (\(M = 10\text{ kg}\) into stationary \(m_3 = 5\text{ kg}\))

Let the incoming speed be \(u = 3\text{ m/s}\). Standard elastic-collision formulas (target at rest):

\[V_M' = \frac{M - m_3}{M + m_3}\,u = \frac{10 - 5}{15}(3) = \frac{5}{15}(3) = 1\text{ m/s}\]
\[V_3' = \frac{2M}{M + m_3}\,u = \frac{20}{15}(3) = 4\text{ m/s}\]

Both move in the original direction (due North): \(M\) at \(1\text{ m/s}\), Puck 3 at \(4\text{ m/s}\).

Part 5 — Verify KE conservation

Before:

\[KE = \tfrac{1}{2}(10)(3)^2 = 45\text{ J}\]

After:

\[KE = \tfrac{1}{2}(10)(1)^2 + \tfrac{1}{2}(5)(4)^2 = 5 + 40 = 45\text{ J}\]
\[45\text{ J} = 45\text{ J} \quad\checkmark\]

Momentum check: before \(= 10(3) = 30\); after \(= 10(1) + 5(4) = 30\).

Feedback & Contributions

Have feedback, want to share your experience, or found a fun variation of this exercise? Email jens.f.mahlmann@dartmouth.edu — we'd love to hear from you!