A person of mass M=90 kg standing on a smooth horizontal plane of ice throws a body of mass m=10 kg horizontally onte same surface. If the distance between the person and body after 10 seconds is 10 metres, the K.E of the person (in Joules) is

Options

A. 0.45
B. 4.5
C. 0.9
D. zero

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
A person of mass \( M = 90 \, \text{kg} \) throws a body of mass \( m = 10 \, \text{kg} \) horizontally on a smooth horizontal plane of ice. The distance between the person and the body after \( t = 10 \, \text{s} \) is \( 10 \, \text{m} \).

### Step 2: Set up the conservation of momentum
Since the ice is smooth (frictionless), the total momentum before and after the throw must be conserved. Initially, both the person and the body are at rest, so the initial momentum is zero. After the throw, we have:
\[ M v_1 + m v_2 = 0 \]
Where:
- \( v_1 \) is the velocity of the person after throwing the body.
- \( v_2 \) is the velocity of the body after being thrown.

### Step 3: Relate the velocities
From the momentum equation, we can express \( v_2 \) in terms of \( v_1 \):
\[ 90 v_1 + 10 v_2 = 0 \implies v_2 = -9 v_1 \]
We only need the magnitudes, so we can write:
\[ v_2 = 9 v_1 \]

### Step 4: Use the distance traveled
The total distance traveled by both the person and the body after \( 10 \, \text{s} \) is \( 10 \, \text{m} \):
\[ s_1 + s_2 = 10 \]
Where:
- \( s_1 = v_1 \cdot t \) (distance traveled by the person)
- \( s_2 = v_2 \cdot t \) (distance traveled by the body)

Substituting the values:
\[ v_1 \cdot 10 + v_2 \cdot 10 = 10 \]
Substituting \( v_2 = 9 v_1 \):
\[ 10 v_1 + 10(9 v_1) = 10 \]
\[ 10 v_1 + 90 v_1 = 10 \implies 100 v_1 = 10 \implies v_1 = \frac{10}{100} = \frac{1}{10} \, \text{m/s} \]

### Step 5: Calculate the kinetic energy of the person
The kinetic energy \( KE \) of the person is given by the formula:
\[ KE = \frac{1}{2} M v_1^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot 90 \cdot \left(\frac{1}{10}\right)^2 \]
\[ KE = \frac{1}{2} \cdot 90 \cdot \frac{1}{100} = \frac{90}{200} = \frac{9}{20} \, \text{J} \]
Converting to decimal:
\[ KE = 0.45 \, \text{J} \]

### Final Answer
The kinetic energy of the person is \( 0.45 \, \text{J} \).

---