A mass `m` moving horizontal (along the x-axis) with velocity `v` collides and sticks to mass of `3m` moving vertically upward (along the y-axis) with velocity `2v`. The final velocity of the combination is

Options

A. `(3)/(2)vhati+(1)/(4)vhatj`
B. `(1)/(4)vhati+(3)/(2)vhatj`
C. `(1)/(3)vhati+(2)/(3)vhatj`
D. `(2)/(3)vhati+(1)/(3)vhatj`

Explanation

To find the final velocity of the combined mass after the collision, we can use the principle of conservation of momentum. Here are the steps to solve the problem:

### Step 1: Identify the masses and their velocities
- Mass \( m_1 = m \) (moving horizontally with velocity \( v \))
- Mass \( m_2 = 3m \) (moving vertically with velocity \( 2v \))

### Step 2: Write down the momentum before the collision
The momentum before the collision can be expressed as the sum of the momenta of both masses:
- Momentum of \( m_1 \) (horizontal):
\[ \vec{p_1} = m \cdot v \hat{i} \]
- Momentum of \( m_2 \) (vertical):
\[ \vec{p_2} = 3m \cdot 2v \hat{j} = 6mv \hat{j} \]

### Step 3: Total momentum before the collision
The total momentum before the collision is:
\[ \vec{p_{initial}} = \vec{p_1} + \vec{p_2} = mv \hat{i} + 6mv \hat{j} \]

### Step 4: Write down the total mass after the collision
After the collision, the two masses stick together, so the total mass \( m_{total} \) is:
\[ m_{total} = m + 3m = 4m \]

### Step 5: Let the final velocity be \( \vec{v_f} \)
Let the final velocity of the combined mass be \( \vec{v_f} \). Then, the momentum after the collision is:
\[ \vec{p_{final}} = m_{total} \cdot \vec{v_f} = 4m \cdot \vec{v_f} \]

### Step 6: Apply conservation of momentum
According to the conservation of momentum:
\[ \vec{p_{initial}} = \vec{p_{final}} \]
Substituting the expressions we derived:
\[ mv \hat{i} + 6mv \hat{j} = 4m \cdot \vec{v_f} \]

### Step 7: Simplify the equation
Dividing through by \( m \):
\[ v \hat{i} + 6v \hat{j} = 4 \vec{v_f} \]
Now, dividing both sides by 4 gives:
\[ \vec{v_f} = \frac{v}{4} \hat{i} + \frac{6v}{4} \hat{j} = \frac{v}{4} \hat{i} + \frac{3v}{2} \hat{j} \]

### Final Result
The final velocity of the combination after the collision is:
\[ \vec{v_f} = \frac{v}{4} \hat{i} + \frac{3v}{2} \hat{j} \]

---