A body of mass `m_(1) ` moving with uniform velocity of 40 m/s collides with another mass `m_(2)` at rest and then the two together begin to moe wit h uniform velocity of 30 m/s. the ratio of their masses `(m_(1))/(m_(2))` is

Options

A. `1:3`
B. `3:1`
C. `1:1.33`
D. `1: 0.75`

Explanation

To solve the problem, we will use the principle of conservation of linear momentum. Let's break down the solution step by step:

### Step 1: Identify the initial and final conditions
- Mass \( m_1 \) is moving with a velocity of \( 40 \, \text{m/s} \).
- Mass \( m_2 \) is at rest, so its initial velocity is \( 0 \, \text{m/s} \).
- After the collision, both masses move together with a velocity of \( 30 \, \text{m/s} \).

### Step 2: Write the expression for initial momentum
The initial momentum \( P_{\text{initial}} \) can be calculated as:
\[ P_{\text{initial}} = m_1 \cdot 40 + m_2 \cdot 0 = 40 m_1 \]

### Step 3: Write the expression for final momentum
The final momentum \( P_{\text{final}} \) after the collision, when both masses move together, is:
\[ P_{\text{final}} = (m_1 + m_2) \cdot 30 \]

### Step 4: Apply the conservation of momentum
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
Substituting the expressions we derived:
\[ 40 m_1 = (m_1 + m_2) \cdot 30 \]

### Step 5: Expand and rearrange the equation
Expanding the right-hand side:
\[ 40 m_1 = 30 m_1 + 30 m_2 \]
Now, rearranging the equation to isolate terms involving \( m_1 \) and \( m_2 \):
\[ 40 m_1 - 30 m_1 = 30 m_2 \]
This simplifies to:
\[ 10 m_1 = 30 m_2 \]

### Step 6: Solve for the ratio of masses
Now, we can find the ratio of \( m_1 \) to \( m_2 \):
\[ \frac{m_1}{m_2} = \frac{30}{10} = 3 \]
Thus, the ratio of their masses \( \frac{m_1}{m_2} \) is:
\[ \frac{m_1}{m_2} = 3:1 \]

### Final Answer
The ratio of the masses \( m_1 \) to \( m_2 \) is \( 3:1 \).

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