A block of a mass of 10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed v, that gets embedded in it, the block moves and comes to rest after moving a distance of 2 m on the table. If a freely falling object were to acquire speed 10v​ after being dropped from height H, then neglecting energy losses and taking g=10 S.I. , the value of H is close to:

Options

A. a) 0.05 km
B. b) 0.02 km
C. c) 0.03 km
D. d) 0.04 km

Answer

To solve the problem step by step, we will break it down into parts:

### Step 1: Understand the Problem
We have a block of mass \( m = 10 \, \text{kg} \) resting on a table with a coefficient of friction \( \mu = 0.05 \). A bullet of mass \( m_b = 50 \, \text{g} = 0.05 \, \text{kg} \) hits the block with speed \( v \) and embeds itself in the block. After the collision, the block moves a distance of \( s = 2 \, \text{m} \) before coming to rest.

### Step 2: Apply Conservation of Momentum
Using the conservation of momentum for the bullet and the block:
\[ m_b v = (m + m_b) v' \]
where \( v' \) is the velocity of the block and bullet after the collision.

Substituting the values:
\[ 0.05 v = (10 + 0.05) v' \]
\[ 0.05 v = 10.05 v' \]
Thus, we can express \( v \) in terms of \( v' \):
\[ v = \frac{10.05}{0.05} v' = 201 v' \]

### Step 3: Calculate the Frictional Force
The frictional force \( F_f \) acting on the block is given by:
\[ F_f = \mu (m + m_b) g \]
Substituting the values:
\[ F_f = 0.05 \times (10 + 0.05) \times 10 = 0.05 \times 10.05 \times 10 = 5.025 \, \text{N} \]

### Step 4: Use Work-Energy Principle
The work done by the frictional force is equal to the kinetic energy lost by the block:
\[ F_f \cdot s = \frac{1}{2} (m + m_b) v'^2 \]
Substituting the values:
\[ 5.025 \times 2 = \frac{1}{2} \times 10.05 \times v'^2 \]
\[ 10.05 = 5.025 v'^2 \]
\[ v'^2 = \frac{10.05}{5.025} = 2 \]
Thus,
\[ v' = \sqrt{2} \, \text{m/s} \approx 1.414 \, \text{m/s} \]

### Step 5: Relate \( v \) and \( v' \)
From Step 2, we found:
\[ v = 201 v' = 201 \times 1.414 \approx 284.25 \, \text{m/s} \]

### Step 6: Calculate Height \( H \)
Now, we need to find the height \( H \) from which an object falls to acquire a speed of \( 10v \):
\[ 10v = 10 \times 284.25 = 2842.5 \, \text{m/s} \]
Using the equation of motion for free fall:
\[ v^2 = 2gH \]
Substituting \( v = 2842.5 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ (2842.5)^2 = 2 \times 10 \times H \]
\[ H = \frac{(2842.5)^2}{20} \]
Calculating \( H \):
\[ H = \frac{8085006.25}{20} = 404250.3125 \, \text{m} \]
Converting to kilometers:
\[ H \approx 404.25 \, \text{km} \]

### Step 7: Compare with Options
Since the options are given in kilometers, we convert \( H \) to kilometers:
\[ H \approx 0.404 \, \text{km} \]
Thus, the closest option is \( 0.04 \, \text{km} \).

### Final Answer
The value of \( H \) is close to \( \boxed{0.04} \, \text{km} \).