A cart is moving along +x direction with a velocityof `4 m//s`. A person on the cart throws a stone with a velocity of `6 m//s` relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of `30^@` with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from the branch of a tree by means of a string of length L. A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine :
(i) The speed of the combined mass immediately after the collision with respect to an observer on the ground,

Explanation

To solve the problem step by step, we will follow the given scenario and apply the principles of physics, particularly the conservation of momentum.

### Step 1: Identify the velocity of the stone relative to the ground
The stone is thrown with a velocity of \(6 \, \text{m/s}\) relative to the cart, which is moving at \(4 \, \text{m/s}\) in the positive x-direction. Since the stone is thrown in the y-z plane, we need to find its velocity components.

1. The vertical component (z-direction) of the stone's velocity can be calculated using the angle of \(30^\circ\):
\[ v_{z} = 6 \cdot \cos(30^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \, \text{m/s} \]

2. The horizontal component (y-direction) of the stone's velocity is:
\[ v_{y} = 6 \cdot \sin(30^\circ) = 6 \cdot \frac{1}{2} = 3 \, \text{m/s} \]

3. The total velocity of the stone in the x-direction (due to the cart's motion) is:
\[ v_{x} = 4 \, \text{m/s} \]

Thus, the velocity of the stone with respect to the ground is:
\[ \vec{v}_{\text{stone}} = (4 \, \text{m/s}, 3 \, \text{m/s}, 3\sqrt{3} \, \text{m/s}) \]

### Step 2: Determine the momentum before the collision
Before the collision, the stone has a momentum given by:
\[ \vec{p}_{\text{stone}} = m \cdot \vec{v}_{\text{stone}} = m \cdot (4, 3, 3\sqrt{3}) \, \text{kg m/s} \]
The object hanging from the tree is at rest, so its momentum is:
\[ \vec{p}_{\text{object}} = 0 \]

### Step 3: Apply conservation of momentum
In a completely inelastic collision, the total momentum before the collision equals the total momentum after the collision. Let \(V\) be the final velocity of the combined mass after the collision.

The total momentum after the collision is:
\[ \vec{p}_{\text{final}} = (m + m) \cdot \vec{V} = 2m \cdot \vec{V} \]

Setting the initial momentum equal to the final momentum:
\[ m \cdot (4, 3, 3\sqrt{3}) = 2m \cdot \vec{V} \]

### Step 4: Solve for the final velocity
Dividing both sides by \(m\) (assuming \(m \neq 0\)):
\[ (4, 3, 3\sqrt{3}) = 2 \cdot \vec{V} \]

Thus, we can find \(\vec{V}\):
\[ \vec{V} = \left( \frac{4}{2}, \frac{3}{2}, \frac{3\sqrt{3}}{2} \right) = (2, 1.5, \frac{3\sqrt{3}}{2}) \, \text{m/s} \]

### Step 5: Calculate the magnitude of the final velocity
To find the speed of the combined mass, we calculate the magnitude of \(\vec{V}\):
\[ V = \sqrt{(2)^2 + (1.5)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} \]
Calculating each term:
\[ V = \sqrt{4 + 2.25 + \frac{27}{4}} = \sqrt{4 + 2.25 + 6.75} = \sqrt{13} \]

Thus, the speed of the combined mass immediately after the collision with respect to an observer on the ground is:
\[ V \approx 3.61 \, \text{m/s} \]