A particle of mass 1.0 g moving with a velocity `vecv_1 = (3hati - 2hatj)` m/s experiences a perfectly inelastic collision with another particle of mass 2.0 g and velocity `vecv_2 = (4hati - 6hatj)` m/s. Find the magnitude of the velocity vector `vecv` of the coalesced particles.

Options

A. 2.3 m/s
B. 5.93 m/s
C. 3.2 m/s
D. 6.4 m/s

Answer

To solve the problem step by step, we will use the principle of conservation of linear momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

### Step 1: Identify the given data
- Mass of particle 1, \( m_1 = 1.0 \, \text{g} = 1.0 \times 10^{-3} \, \text{kg} \)
- Velocity of particle 1, \( \vec{v_1} = (3 \hat{i} - 2 \hat{j}) \, \text{m/s} \)
- Mass of particle 2, \( m_2 = 2.0 \, \text{g} = 2.0 \times 10^{-3} \, \text{kg} \)
- Velocity of particle 2, \( \vec{v_2} = (4 \hat{i} - 6 \hat{j}) \, \text{m/s} \)

### Step 2: Calculate the initial momentum of each particle
The momentum of a particle is given by the product of its mass and velocity.

- Momentum of particle 1:
\[ \vec{p_1} = m_1 \vec{v_1} = (1.0 \times 10^{-3} \, \text{kg}) (3 \hat{i} - 2 \hat{j}) = (3 \times 10^{-3} \hat{i} - 2 \times 10^{-3} \hat{j}) \, \text{kg m/s} \]

- Momentum of particle 2:
\[ \vec{p_2} = m_2 \vec{v_2} = (2.0 \times 10^{-3} \, \text{kg}) (4 \hat{i} - 6 \hat{j}) = (8 \times 10^{-3} \hat{i} - 12 \times 10^{-3} \hat{j}) \, \text{kg m/s} \]

### Step 3: Calculate the total initial momentum
Now, we can find the total initial momentum by adding the momenta of both particles:
\[ \vec{p_{\text{initial}}} = \vec{p_1} + \vec{p_2} = (3 \times 10^{-3} \hat{i} - 2 \times 10^{-3} \hat{j}) + (8 \times 10^{-3} \hat{i} - 12 \times 10^{-3} \hat{j}) \]
\[ \vec{p_{\text{initial}}} = (3 + 8) \times 10^{-3} \hat{i} + (-2 - 12) \times 10^{-3} \hat{j} = 11 \times 10^{-3} \hat{i} - 14 \times 10^{-3} \hat{j} \, \text{kg m/s} \]

### Step 4: Calculate the total mass after the collision
Since the collision is perfectly inelastic, the two particles stick together. Therefore, the total mass after the collision is:
\[ m_{\text{total}} = m_1 + m_2 = (1.0 \times 10^{-3} + 2.0 \times 10^{-3}) \, \text{kg} = 3.0 \times 10^{-3} \, \text{kg} \]

### Step 5: Use conservation of momentum to find the final velocity
According to the conservation of momentum:
\[ \vec{p_{\text{final}}} = \vec{p_{\text{initial}}} \]
Let \( \vec{v} \) be the final velocity of the coalesced particles. Then:
\[ m_{\text{total}} \vec{v} = \vec{p_{\text{initial}}} \]
\[ (3.0 \times 10^{-3}) \vec{v} = (11 \times 10^{-3} \hat{i} - 14 \times 10^{-3} \hat{j}) \]
Dividing both sides by \( 3.0 \times 10^{-3} \):
\[ \vec{v} = \frac{11 \times 10^{-3}}{3.0 \times 10^{-3}} \hat{i} - \frac{14 \times 10^{-3}}{3.0 \times 10^{-3}} \hat{j} = \frac{11}{3} \hat{i} - \frac{14}{3} \hat{j} \, \text{m/s} \]

### Step 6: Calculate the magnitude of the velocity vector
The magnitude of the velocity vector \( \vec{v} \) is given by:
\[ |\vec{v}| = \sqrt{\left(\frac{11}{3}\right)^2 + \left(-\frac{14}{3}\right)^2} \]
Calculating the squares:
\[ |\vec{v}| = \sqrt{\frac{121}{9} + \frac{196}{9}} = \sqrt{\frac{317}{9}} = \frac{\sqrt{317}}{3} \]
Calculating the square root:
\[ \sqrt{317} \approx 17.8 \implies |\vec{v}| \approx \frac{17.8}{3} \approx 5.93 \, \text{m/s} \]

### Final Answer
The magnitude of the velocity vector \( \vec{v} \) of the coalesced particles is approximately \( 5.93 \, \text{m/s} \).

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