A particle dropped from a height of 100 m, collides with the horizontal surface where coefficient of restitution is 0.5, the velocity attained by the particle in `5^(th)` collision is

Options

A. 4.2 m/s
B. 6.2 m/s
C. 8.4 m/s
D. 6.4 m/s

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the initial velocity just before the first collision
The particle is dropped from a height \( h = 100 \, \text{m} \). We can use the third equation of motion to find the velocity just before the collision:

\[ v = \sqrt{u^2 + 2gh} \]

Where:
- \( u = 0 \) (initial velocity, since the particle is dropped),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 100 \, \text{m} \) (height).

Substituting the values:

\[ v = \sqrt{0 + 2 \times 10 \times 100} = \sqrt{2000} = 44.72 \, \text{m/s} \]

### Step 2: Calculate the velocity after the first collision
The coefficient of restitution \( e = 0.5 \). The velocity after the first collision is given by:

\[ v_1 = e \cdot v \]

Substituting the values:

\[ v_1 = 0.5 \times 44.72 = 22.36 \, \text{m/s} \]

### Step 3: Calculate the velocity after subsequent collisions
The velocity after each collision can be expressed as:

- After the second collision: \( v_2 = e \cdot v_1 = e^2 \cdot v \)
- After the third collision: \( v_3 = e \cdot v_2 = e^3 \cdot v \)
- After the fourth collision: \( v_4 = e \cdot v_3 = e^4 \cdot v \)
- After the fifth collision: \( v_5 = e \cdot v_4 = e^5 \cdot v \)

Thus, the velocity after the fifth collision is:

\[ v_5 = e^5 \cdot v \]

### Step 4: Substitute the values to find the velocity after the fifth collision
Now substituting the values we have:

\[ v_5 = (0.5)^5 \cdot 44.72 \]

Calculating \( (0.5)^5 = \frac{1}{32} = 0.03125 \):

\[ v_5 = 0.03125 \cdot 44.72 \approx 1.4 \, \text{m/s} \]

### Final Answer
The velocity attained by the particle in the 5th collision is approximately \( 1.4 \, \text{m/s} \).

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