A boy of mass `60kg` is standing over a platform of mass `40kg` placed over a smooth horizontal surface. He throws a stone of mass `1kg` with velocity `v=10m//s` at an angle of `45^@` with respect to the ground. Find the displacement of the platform (with boy) on the horizontal surface when the stone lands on the ground. Take `g=10m//s^2`.

Explanation

To solve the problem, we need to find the displacement of the platform (with the boy) on the horizontal surface when the stone lands on the ground. We will follow these steps:

### Step 1: Calculate the Range of the Stone
The stone is thrown with a velocity \( v = 10 \, \text{m/s} \) at an angle \( \theta = 45^\circ \). The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

Substituting the values:
\[ R = \frac{(10)^2 \sin(90^\circ)}{10} = \frac{100 \cdot 1}{10} = 10 \, \text{m} \]

### Step 2: Determine the Initial Momentum of the System
Before the stone is thrown, the system (boy + platform) is at rest. Therefore, the initial momentum of the system is:
\[ p_{\text{initial}} = 0 \]

### Step 3: Calculate the Final Momentum of the Stone
When the stone is thrown, it has a mass \( m_s = 1 \, \text{kg} \) and a velocity \( v_s = 10 \, \text{m/s} \) at \( 45^\circ \). The horizontal component of the stone's velocity is:
\[ v_{sx} = v_s \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s} \]

The momentum of the stone in the horizontal direction is:
\[ p_s = m_s v_{sx} = 1 \cdot 5\sqrt{2} = 5\sqrt{2} \, \text{kg m/s} \]

### Step 4: Apply Conservation of Momentum
By conservation of momentum, the momentum of the system after the stone is thrown must equal the initial momentum:
\[ p_{\text{initial}} = p_{\text{final}} \]
Let \( v_p \) be the velocity of the platform (with the boy) after the stone is thrown. The total mass of the platform and the boy is \( m_b + m_p = 60 + 40 = 100 \, \text{kg} \).

Thus, we have:
\[ 0 = p_s + p_p \implies 0 = 5\sqrt{2} + 100 v_p \]
Solving for \( v_p \):
\[ 100 v_p = -5\sqrt{2} \implies v_p = -\frac{5\sqrt{2}}{100} = -\frac{\sqrt{2}}{20} \, \text{m/s} \]

### Step 5: Calculate the Time of Flight of the Stone
The time of flight \( t \) for the stone can be calculated using the vertical motion:
\[ t = \frac{2u \sin(\theta)}{g} = \frac{2 \cdot 10 \cdot \frac{1}{\sqrt{2}}}{10} = \frac{2\sqrt{2}}{2} = \sqrt{2} \, \text{s} \]

### Step 6: Calculate the Displacement of the Platform
The displacement \( d \) of the platform during the time the stone is in the air is given by:
\[ d = v_p \cdot t = -\frac{\sqrt{2}}{20} \cdot \sqrt{2} = -\frac{2}{20} = -0.1 \, \text{m} = -10 \, \text{cm} \]

### Final Answer
The displacement of the platform (with the boy) on the horizontal surface when the stone lands on the ground is:
\[ \text{Displacement} = -0.1 \, \text{m} \text{ (or -10 cm)} \]