Two bodies having masses in the ratio 2:3 fall freely under gravity from heights 9:16. The ratio of their linear momenta on touching, the ground is

Options

A. `2:9`
B. `3:16`
C. `1:2`
D. `3:2`

Explanation

To solve the problem step by step, we will find the ratio of linear momenta of two bodies falling freely under gravity from different heights.

### Step 1: Understand the relationship between momentum, mass, and velocity
The momentum (P) of an object is given by the formula:
\[ P = m \cdot v \]
where \( m \) is the mass and \( v \) is the velocity of the object.

### Step 2: Determine the final velocity of the objects just before they touch the ground
Since both bodies are falling freely under gravity, we can use the kinematic equation:
\[ v^2 = u^2 + 2gh \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is 0 for free fall),
- \( g \) is the acceleration due to gravity,
- \( h \) is the height from which the body falls.

Given that \( u = 0 \), the equation simplifies to:
\[ v = \sqrt{2gh} \]

### Step 3: Calculate the momenta of both bodies
Let:
- Mass of body 1, \( m_1 \), be \( 2k \) (where \( k \) is a constant),
- Mass of body 2, \( m_2 \), be \( 3k \),
- Height from which body 1 falls, \( h_1 = 9 \),
- Height from which body 2 falls, \( h_2 = 16 \).

The final velocities for both bodies just before they touch the ground will be:
- For body 1:
\[ v_1 = \sqrt{2g \cdot h_1} = \sqrt{2g \cdot 9} = \sqrt{18g} \]
- For body 2:
\[ v_2 = \sqrt{2g \cdot h_2} = \sqrt{2g \cdot 16} = \sqrt{32g} \]

### Step 4: Write the expressions for the momenta
Now we can write the momenta for both bodies:
- Momentum of body 1:
\[ P_1 = m_1 \cdot v_1 = (2k) \cdot \sqrt{18g} = 2k \sqrt{18g} \]
- Momentum of body 2:
\[ P_2 = m_2 \cdot v_2 = (3k) \cdot \sqrt{32g} = 3k \sqrt{32g} \]

### Step 5: Find the ratio of the momenta
Now we can find the ratio of their momenta:
\[ \frac{P_1}{P_2} = \frac{2k \sqrt{18g}}{3k \sqrt{32g}} \]

The \( k \) and \( g \) cancel out:
\[ \frac{P_1}{P_2} = \frac{2 \sqrt{18}}{3 \sqrt{32}} \]

### Step 6: Simplify the ratio
Now we simplify the ratio:
\[ \frac{P_1}{P_2} = \frac{2 \cdot \sqrt{18}}{3 \cdot \sqrt{32}} = \frac{2 \cdot 3\sqrt{2}}{3 \cdot 4\sqrt{2}} \]
\[ = \frac{2 \cdot 3}{3 \cdot 4} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer
The ratio of their linear momenta on touching the ground is:
\[ \frac{P_1}{P_2} = 1:2 \]