Two blocks A and B of masses 10 kg and 5 kg respectively are placed inside a cart which is moving horizontally with an acceleration of`2 m/s^2` as shown in the figure. If all the surfaces are smooth then the ratio of normal reaction between the block A and vertical surface of the cart and normal reaction between the blocks A and B is

Options

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

Answer

To solve the problem, we need to find the ratio of the normal reaction between block A and the vertical surface of the cart (let's call it \( R_2 \)) to the normal reaction between blocks A and B (let's call it \( R_1 \)).

### Step-by-Step Solution:

1. **Identify the Forces Acting on Block B:**
- Block B has a mass \( m_B = 5 \, \text{kg} \).
- The cart is accelerating with \( a = 2 \, \text{m/s}^2 \).
- The normal reaction force \( R_1 \) acts on block B from block A.

Using Newton's second law, the force acting on block B can be expressed as:
\[ R_1 = m_B \cdot a \]
Substituting the values:
\[ R_1 = 5 \, \text{kg} \cdot 2 \, \text{m/s}^2 = 10 \, \text{N} \]

2. **Identify the Forces Acting on Block A:**
- Block A has a mass \( m_A = 10 \, \text{kg} \).
- The normal reaction force \( R_2 \) acts on block A from the vertical surface of the cart and also includes the force due to block B.

The total force acting on block A can be expressed as:
\[ R_2 = R_1 + m_A \cdot a \]
Substituting the values:
\[ R_2 = 10 \, \text{N} + (10 \, \text{kg} \cdot 2 \, \text{m/s}^2) = 10 \, \text{N} + 20 \, \text{N} = 30 \, \text{N} \]

3. **Calculate the Ratio of the Normal Reactions:**
Now we can find the ratio of \( R_2 \) to \( R_1 \):
\[ \text{Ratio} = \frac{R_2}{R_1} = \frac{30 \, \text{N}}{10 \, \text{N}} = 3 \]

4. **Final Result:**
The ratio of the normal reaction between block A and the vertical surface of the cart to the normal reaction between blocks A and B is:
\[ \text{Ratio} = 3:1 \]