A bullet weighing 10 g and moving with a velocity 300 m/s strikes a 5 kg block of ice and drop dead. The ice block is kept on smooth surface. The speed of the block after the the collision is

Options

A. 6 cm/s
B. 60 cm/s
C. 6 m/s
D. 0.6 cm/s

Answer

To solve the problem, we will use the principle of conservation of linear momentum. Here are the steps to find the speed of the block after the collision:

### Step 1: Understand the Problem
We have a bullet weighing 10 g (0.01 kg) moving at a velocity of 300 m/s that strikes a block of ice weighing 5 kg. After the collision, the bullet embeds itself in the block, and we need to find the speed of the block after the collision.

### Step 2: Write Down the Given Data
- Mass of the bullet (m_bullet) = 10 g = 0.01 kg
- Velocity of the bullet (v_bullet) = 300 m/s
- Mass of the block (m_block) = 5 kg
- Initial velocity of the block (v_block_initial) = 0 m/s (since it is at rest)

### Step 3: Apply the Conservation of Momentum
According to the law of conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]

Initial momentum before the collision:
\[ p_{initial} = m_{bullet} \cdot v_{bullet} + m_{block} \cdot v_{block\_initial} \]
\[ p_{initial} = (0.01 \, \text{kg}) \cdot (300 \, \text{m/s}) + (5 \, \text{kg}) \cdot (0 \, \text{m/s}) \]
\[ p_{initial} = 3 \, \text{kg m/s} + 0 \]
\[ p_{initial} = 3 \, \text{kg m/s} \]

Final momentum after the collision:
Let \( V \) be the final velocity of the block and bullet together.
\[ p_{final} = (m_{bullet} + m_{block}) \cdot V \]
\[ p_{final} = (0.01 \, \text{kg} + 5 \, \text{kg}) \cdot V \]
\[ p_{final} = (5.01 \, \text{kg}) \cdot V \]

### Step 4: Set Initial Momentum Equal to Final Momentum
\[ 3 \, \text{kg m/s} = (5.01 \, \text{kg}) \cdot V \]

### Step 5: Solve for V
\[ V = \frac{3 \, \text{kg m/s}}{5.01 \, \text{kg}} \]
\[ V \approx 0.5984 \, \text{m/s} \]

### Step 6: Convert to cm/s
To convert from m/s to cm/s:
\[ V \approx 0.5984 \, \text{m/s} \times 100 \, \text{cm/m} \]
\[ V \approx 59.84 \, \text{cm/s} \]

### Final Answer
The speed of the block after the collision is approximately **60 cm/s**.

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