An object A of mass `1 kg` is projected vertically upward with a speed of `20 m//s`. At the same moment another object B of mass `3 kg`, which in intially above the object A, is dropped from a height `h = 20 m`. The two point like objects (A and B) collision and stick to each other. The kinetic energy is K (in J) of the combined mass after collision, find the value of `K//25`.

Explanation

To solve the problem step by step, we will analyze the motion of both objects A and B, calculate their velocities just before the collision, and then find the kinetic energy of the combined mass after the collision.

### Step 1: Determine the time of collision
Object A is projected upward with an initial speed of \(20 \, \text{m/s}\) and object B is dropped from a height of \(20 \, \text{m}\). We need to find the time \(t\) it takes for them to collide.

Using the relative motion concept:
- The relative displacement between A and B is \(20 \, \text{m}\).
- The relative velocity of A with respect to B is \(20 \, \text{m/s}\) (upward).

Using the formula for relative motion:
\[ \text{Relative Displacement} = \text{Relative Velocity} \times t \]
\[ 20 = 20 \times t \]
\[ t = 1 \, \text{s} \]

### Step 2: Calculate the velocity of object A just before the collision
Using the equation of motion for object A:
\[ V_A = U_A + a_A \cdot t \]
Where:
- \(U_A = 20 \, \text{m/s}\) (initial velocity of A)
- \(a_A = -10 \, \text{m/s}^2\) (acceleration due to gravity acting downward)
- \(t = 1 \, \text{s}\)

Calculating \(V_A\):
\[ V_A = 20 + (-10) \cdot 1 = 20 - 10 = 10 \, \text{m/s} \]

### Step 3: Calculate the velocity of object B just before the collision
Using the equation of motion for object B:
\[ V_B = U_B + a_B \cdot t \]
Where:
- \(U_B = 0 \, \text{m/s}\) (initial velocity of B)
- \(a_B = -10 \, \text{m/s}^2\) (acceleration due to gravity acting downward)
- \(t = 1 \, \text{s}\)

Calculating \(V_B\):
\[ V_B = 0 + (-10) \cdot 1 = -10 \, \text{m/s} \]

### Step 4: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision.

Before the collision:
\[ \text{Total momentum} = m_A \cdot V_A + m_B \cdot V_B \]
Where:
- \(m_A = 1 \, \text{kg}\)
- \(m_B = 3 \, \text{kg}\)

Calculating total momentum:
\[ \text{Total momentum} = 1 \cdot 10 + 3 \cdot (-10) = 10 - 30 = -20 \, \text{kg m/s} \]

After the collision, the combined mass is \(m_A + m_B = 4 \, \text{kg}\) and let the final velocity be \(V_f\):
\[ \text{Total momentum after collision} = (m_A + m_B) \cdot V_f = 4 \cdot V_f \]
Setting the total momentum before and after equal:
\[ -20 = 4 \cdot V_f \]
Solving for \(V_f\):
\[ V_f = \frac{-20}{4} = -5 \, \text{m/s} \]

### Step 5: Calculate the kinetic energy of the combined mass after the collision
The kinetic energy \(K\) of the combined mass is given by:
\[ K = \frac{1}{2} (m_A + m_B) V_f^2 \]
Substituting the values:
\[ K = \frac{1}{2} \cdot 4 \cdot (-5)^2 = \frac{1}{2} \cdot 4 \cdot 25 = 2 \cdot 25 = 50 \, \text{J} \]

### Step 6: Find the value of \(K/25\)
\[ \frac{K}{25} = \frac{50}{25} = 2 \]

### Final Answer
The value of \(K/25\) is \(2\).

---