A rock mass of 3kg was in rest. Due to an explosion the rock devided onto 1:1:3 ratio the both same mass rock got the velocity of 18m/s in 180° of each other now what is the velocity of big pile of rock



To solve this problem, we can apply the **principle of conservation of momentum**. According to this principle, the total momentum of a system before an event (like an explosion) is equal to the total momentum after the event, provided no external forces act on the system.

### Steps:

1. **Before the explosion**:
   - The rock was at rest, so the total initial momentum was zero.

2. **After the explosion**:
   - The rock divided into three parts with a mass ratio of \(1:1:3\).
   - Two of the smaller parts have the same mass, and each moves with a velocity of \(18 \, \text{m/s}\) in opposite directions (at \(180^\circ\) to each other).
   - Let's denote the mass of the larger piece as \(m_3\) and its velocity as \(v_3\). We need to find \(v_3\).

#### Masses:
- Let the mass of the rock before the explosion be \(3 \, \text{kg}\).
- If the mass ratio is \(1:1:3\), we can divide the total mass \(3 \, \text{kg}\) accordingly:
  - The mass of each smaller piece: \(m_1 = m_2 = \frac{3 \, \text{kg}}{5} = 0.6 \, \text{kg}\).
  - The mass of the larger piece: \(m_3 = 3 \times 0.6 \, \text{kg} = 1.8 \, \text{kg}\).

#### Momentum of the two smaller rocks:
- The momentum of each smaller rock is given by:
  \[
  p_1 = p_2 = m_1 \times v_1 = 0.6 \times 18 = 10.8 \, \text{kg} \cdot \text{m/s}
  \]
- Since these rocks move in opposite directions, their momenta cancel each other out, i.e., the total momentum of the smaller rocks is:
  \[
  p_{\text{small rocks}} = 10.8 - 10.8 = 0 \, \text{kg} \cdot \text{m/s}
  \]

#### Momentum of the larger piece:
- To conserve momentum, the total momentum of the system must remain zero. Therefore, the momentum of the larger piece of rock must also be zero relative to the other pieces. 
- Let \(v_3\) be the velocity of the larger piece:
  \[
  p_3 = m_3 \times v_3 = 1.8 \times v_3
  \]
- The total momentum after the explosion is:
  \[
  p_{\text{total}} = p_{\text{small rocks}} + p_3 = 0 + 1.8 \times v_3
  \]

Since the total initial momentum was zero, the total momentum after the explosion must also be zero:
\[
1.8 \times v_3 = 0
\]

Thus, the velocity of the larger rock is \(v_3 = 0 \, \text{m/s}\).

### Conclusion:
The velocity of the larger piece of rock after the explosion is \(0 \, \text{m/s}\). 

Would you like further explanation or details on how momentum works?

Here are 5 related questions to explore further:
1. What is the law of conservation of momentum?
2. How does the direction of velocity affect momentum?
3. What happens when more than two objects are involved in a momentum conservation problem?
4. How would the result change if the masses were in a different ratio?
5. How is kinetic energy affected in inelastic explosions like this?

**Tip:** In problems involving explosions or collisions, always check if the momentum is conserved along all directions (e.g., both horizontal and vertical).

A rock mass of 3kg was in rest. Due to an explosion the rock divided onto 1:1:3 ratio. The two same mass rocks got the velocity of 18m/s in 180° of each other. What is the velocity of the big pile of rock?

Learn how to solve a momentum conservation problem involving a 3kg rock that splits into three parts in a 1:1:3 ratio. Discover how the velocities of the smaller rocks affect the velocity of the larger rock, using the principle of conservation of momentum. Suitable for Grades 10-12.

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