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.)

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:
\[
PE_{\text{spring}} = \frac{1}{2} k x^2
\]
where:
- \( k = 260 \, \text{N/m} \) (force constant),
- \( x = 9.2 \, \text{cm} = 0.092 \, \text{m} \) (compression of the spring).

\[
PE_{\text{spring}} = \frac{1}{2} \times 260 \times (0.092)^2
\]
Calculating this gives:
\[
PE_{\text{spring}} = 1.10088 \, \text{J}
\]

### 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 \( v_1 \) be the final velocity of \( m_1 \) and \( v_2 \) the final velocity of \( m_2 \). Conservation of momentum tells us:
\[
m_1 v_1 + m_2 v_2 = 0
\]
which implies:
\[
v_2 = -\frac{m_1}{m_2} v_1
\]

### Step 3: Conservation of Energy in Terms of Velocities
The total kinetic energy after release equals the potential energy initially stored in the spring:
\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = PE_{\text{spring}}
\]

Substitute \( v_2 = -\frac{m_1}{m_2} v_1 \) into the energy equation:
\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \left( -\frac{m_1}{m_2} v_1 \right)^2 = PE_{\text{spring}}
\]

Simplifying:
\[
\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 \frac{m_1^2}{m_2^2} v_1^2 = PE_{\text{spring}}
\]
\[
\frac{1}{2} v_1^2 \left( m_1 + \frac{m_1^2}{m_2} \right) = PE_{\text{spring}}
\]

Solve for \( v_1 \):
\[
v_1^2 = \frac{2 \times PE_{\text{spring}}}{m_1 + \frac{m_1^2}{m_2}}
\]
\[
v_1 = \sqrt{\frac{2 \times 1.10088}{0.42 + \frac{(0.42)^2}{0.94}}}
\]

Let's calculate \( v_1 \) and \( v_2 = -\frac{m_1}{m_2} v_1 \).The final velocities of the two objects are:

- \( v_1 = 1.90 \, \text{m/s} \) (to the right),
- \( v_2 = -0.85 \, \text{m/s} \) (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.

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. 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.

Solve for the final velocities of two masses (0.42 kg and 0.94 kg) on a frictionless surface released by a compressed spring with a force constant of 260 N/m. This problem covers conservation of energy and momentum principles. Suitable for high school physics students.

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