Math Problem Statement

Two objects of masses m1 = 0.42 kg and m2 = 0.94 kg are placed on a horizontal frictionless surface and a compressed spring of force constant k = 260 N/m is placed between them as in figure (a). Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 9.2 cm. If the objects are released from rest, find the final velocity of each object as shown in figure (b). (Let the positive direction be to the right. Indicate the direction with the sign of your answer.)

Solution

To solve this problem, we can use the concepts of conservation of energy and conservation of momentum.

Step 1: Conservation of Energy

The potential energy stored in the compressed spring will convert into the kinetic energy of both objects since there's no external force or friction.

The initial potential energy of the spring: PEspring=12kx2PE_{\text{spring}} = \frac{1}{2} k x^2 where:

  • k=260N/mk = 260 \, \text{N/m} (force constant),
  • x=9.2cm=0.092mx = 9.2 \, \text{cm} = 0.092 \, \text{m} (compression of the spring).

PEspring=12×260×(0.092)2PE_{\text{spring}} = \frac{1}{2} \times 260 \times (0.092)^2 Calculating this gives: PEspring=1.10088JPE_{\text{spring}} = 1.10088 \, \text{J}

Step 2: Conservation of Momentum

Since no external force acts on the system, the total momentum is conserved. Initially, the objects are at rest, so the total initial momentum is zero.

Let v1v_1 be the final velocity of m1m_1 and v2v_2 the final velocity of m2m_2. Conservation of momentum tells us: m1v1+m2v2=0m_1 v_1 + m_2 v_2 = 0 which implies: v2=m1m2v1v_2 = -\frac{m_1}{m_2} v_1

Step 3: Conservation of Energy in Terms of Velocities

The total kinetic energy after release equals the potential energy initially stored in the spring: 12m1v12+12m2v22=PEspring\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = PE_{\text{spring}}

Substitute v2=m1m2v1v_2 = -\frac{m_1}{m_2} v_1 into the energy equation: 12m1v12+12m2(m1m2v1)2=PEspring\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \left( -\frac{m_1}{m_2} v_1 \right)^2 = PE_{\text{spring}}

Simplifying: 12m1v12+12m2m12m22v12=PEspring\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \frac{m_1^2}{m_2^2} v_1^2 = PE_{\text{spring}} 12v12(m1+m12m2)=PEspring\frac{1}{2} v_1^2 \left( m_1 + \frac{m_1^2}{m_2} \right) = PE_{\text{spring}}

Solve for v1v_1: v12=2×PEspringm1+m12m2v_1^2 = \frac{2 \times PE_{\text{spring}}}{m_1 + \frac{m_1^2}{m_2}} v1=2×1.100880.42+(0.42)20.94v_1 = \sqrt{\frac{2 \times 1.10088}{0.42 + \frac{(0.42)^2}{0.94}}}

Let's calculate v1v_1 and v2=m1m2v1v_2 = -\frac{m_1}{m_2} v_1.The final velocities of the two objects are:

  • v1=1.90m/sv_1 = 1.90 \, \text{m/s} (to the right),
  • v2=0.85m/sv_2 = -0.85 \, \text{m/s} (to the left).

These results account for both conservation of momentum and energy.

Would you like further details on any step, or have any questions?

Here are five related questions for deeper understanding:

  1. What would change if there were friction on the surface?
  2. How would the final velocities differ if the spring were attached to one of the objects?
  3. What effect would a stronger or weaker spring constant have on the final speeds?
  4. How is momentum conserved in systems where objects move in opposite directions?
  5. How can this setup demonstrate the principles of an idealized elastic collision?

Tip: Always start with conservation laws in physics problems involving no external forces, as they simplify analysis.

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Math Problem Analysis

Mathematical Concepts

Physics
Conservation of Energy
Conservation of Momentum

Formulas

Potential Energy of Spring: PE_spring = (1/2) k x^2
Conservation of Momentum: m1 * v1 + m2 * v2 = 0
Kinetic Energy: KE = (1/2) m * v^2

Theorems

Conservation of Momentum
Conservation of Energy

Suitable Grade Level

Grade 11-12