Two bodies of masses 0.1 kg and 0.4 kg move towards each other with the velocities 1 m/s and 0.1 m/s respectively, After collision they stick together. In 10 sec the combined mass travels

Options

A. 120 m
B. 0.12 m
C. 12 m
D. 1.2 m

Answer

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify the masses and velocities of the two bodies.
- Mass of body 1 (m1) = 0.1 kg, Velocity of body 1 (v1) = 1 m/s (towards the right).
- Mass of body 2 (m2) = 0.4 kg, Velocity of body 2 (v2) = 0.1 m/s (towards the left).

### Step 2: Calculate the initial momentum of both bodies.
- The momentum of body 1 (p1) = m1 * v1 = 0.1 kg * 1 m/s = 0.1 kg·m/s (to the right).
- The momentum of body 2 (p2) = m2 * v2 = 0.4 kg * (-0.1 m/s) = -0.04 kg·m/s (to the left).

### Step 3: Calculate the total initial momentum of the system.
- Total initial momentum (P_initial) = p1 + p2 = 0.1 kg·m/s - 0.04 kg·m/s = 0.06 kg·m/s.

### Step 4: Apply the conservation of momentum principle.
- After the collision, the two bodies stick together, so their combined mass (M) = m1 + m2 = 0.1 kg + 0.4 kg = 0.5 kg.
- Let the final velocity after collision be V.
- According to the conservation of momentum: P_initial = P_final
- Therefore, 0.06 kg·m/s = M * V = 0.5 kg * V.

### Step 5: Solve for the final velocity (V).
- Rearranging gives: V = P_initial / M = 0.06 kg·m/s / 0.5 kg = 0.12 m/s.

### Step 6: Calculate the distance traveled in 10 seconds.
- The distance (d) traveled by the combined mass after the collision can be calculated using the formula:
\[ d = V \times t \]
- Where V = 0.12 m/s and t = 10 seconds.
- Therefore, d = 0.12 m/s * 10 s = 1.2 meters.

### Final Answer:
The combined mass travels a distance of **1.2 meters** in 10 seconds.

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