Nuclear particles travel at tremendous speeds. Two such particles d and β particles mere observed to be traveling at 10³ mm/sec and 3.2×10³ mm/sec and are approaching each other. Upon impact, the two particles give off heat energy proportional to their initial distance and the square of the time to impact if the initial distance between the two particles is 1 mm, how long will it be before they collide? How much heat energy will be given off upon impact?

To solve this problem, we will first find the time until the two particles collide and then compute the heat energy given off upon impact.

### Step 1: Finding the Time to Collision

Let:
- The speed of the first particle (d particle) be \( v_1 = 10^3 \, \text{mm/sec} \).
- The speed of the second particle (β particle) be \( v_2 = 3.2 \times 10^3 \, \text{mm/sec} \).
- The initial distance between the two particles be \( D = 1 \, \text{mm} \).

Since the particles are approaching each other, the relative speed between them is the sum of their individual speeds:
\[
v_{\text{relative}} = v_1 + v_2 = 10^3 + 3.2 \times 10^3 = 4.2 \times 10^3 \, \text{mm/sec}.
\]

The time to collision, \( t \), is given by the formula:
\[
t = \frac{\text{initial distance}}{\text{relative speed}} = \frac{D}{v_{\text{relative}}}.
\]

Substitute the given values:
\[
t = \frac{1 \, \text{mm}}{4.2 \times 10^3 \, \text{mm/sec}}.
\]

Calculating this gives:
\[
t \approx \frac{1}{4200} \, \text{sec}.
\]

Now, let's compute the exact value.

### Step 2: Calculating the Heat Energy Given Off Upon Impact

The heat energy \( E \) given off upon impact is proportional to the initial distance \( D \) and the square of the time to impact \( t^2 \). Let the constant of proportionality be \( k \). Therefore, the heat energy \( E \) can be expressed as:
\[
E = k \cdot D \cdot t^2.
\]

We have \( D = 1 \, \text{mm} \), and we will use the value of \( t \) we just computed. Substituting these values, we get:
\[
E = k \cdot 1 \cdot \left( \frac{1}{4.2 \times 10^3} \right)^2.
\]

Let's compute both the time to collision and the expression for energy.1. **Time to Collision:**
   The time before the two particles collide is approximately \( t \approx 0.000238 \) seconds.

2. **Heat Energy Given Off Upon Impact:**
   The heat energy \( E \) given off upon impact, in terms of the proportional constant \( k \), is:
   \[
   E = k \times 5.67 \times 10^{-8}.
   \]

This means that the heat energy depends on the proportional constant \( k \). To find the exact value, the value of \( k \) would need to be specified.

Would you like more details or have any questions?

### Related Questions:
1. How is relative speed calculated when two objects are moving towards each other?
2. What happens to the time to collision if the speeds of the particles are doubled?
3. How would the formula for heat energy change if the distance were tripled?
4. What units should be used when calculating heat energy?
5. Can we express the relationship between speed, distance, and time graphically?

**Tip:** Remember to always check units when working with physical quantities to ensure consistency in your calculations.

Explore the collision of nuclear particles d and β traveling at 10³ mm/sec and 3.2×10³ mm/sec. Learn how to calculate the time before they collide and the heat energy given off upon impact using principles of kinematics and proportionality. Suitable for students in Grades 11-12.

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