A ball falls from a height such that it strikes the floor of lift at 10 m/s. If lift is moving in the upward direction with a velocity 1 m/s, then velocity with which the ball rebounds after elastic collision will be

Options

A. 11 m/s
B. 12 m/s
C. 13 m/s
D. 9 m/s

Answer

To solve the problem, we need to analyze the motion of the ball and the lift during the collision.

**Step 1: Understand the initial conditions.**
- The ball is falling and strikes the floor of the lift with a velocity of 10 m/s downward.
- The lift is moving upward with a velocity of 1 m/s.

**Step 2: Determine the relative velocity of approach.**
- The relative velocity of the ball with respect to the lift just before the collision can be calculated as:
\[ \text{Relative velocity of approach} = \text{Velocity of ball} + \text{Velocity of lift} = 10 \, \text{m/s} + 1 \, \text{m/s} = 11 \, \text{m/s} \]
(Note: Both velocities are in opposite directions, hence they are added.)

**Step 3: Apply the principle of elastic collision.**
- In an elastic collision, the relative velocity of separation after the collision is equal to the relative velocity of approach before the collision.
- Let \( v \) be the velocity of the ball after the collision. The lift continues to move upward at 1 m/s.

**Step 4: Determine the relative velocity of separation.**
- The relative velocity of separation can be expressed as:
\[ \text{Relative velocity of separation} = \text{Velocity of ball after collision} - \text{Velocity of lift} = v - 1 \, \text{m/s} \]

**Step 5: Set up the equation using the elastic collision condition.**
- According to the principle of elastic collision:
\[ \text{Relative velocity of separation} = \text{Relative velocity of approach} \]
Thus, we have:
\[ v - 1 = 11 \]

**Step 6: Solve for \( v \).**
- Rearranging the equation gives:
\[ v = 11 + 1 = 12 \, \text{m/s} \]

**Conclusion:**
- The velocity with which the ball rebounds after the elastic collision is **12 m/s** upward.

---