Question

A body of mass \(5\) kg collides with a wall with a speed of \(50\,\text{m s}^{-1}\) and rebounds with the same speed. If the time of contact of the body with the wall is \(\frac{1}{20}\) s, the force exerted on the wall is

• A. \(0.5\times 10^{4}\,\text{N}\)
• B. \(2.5\times 10^{4}\,\text{N}\)
• C. \(2\times 10^{3}\,\text{N}\)
• D. \(1\times 10^{4}\,\text{N}\)
• E. \(4\times 10^{3}\,\text{N}\)

Hint:

When a body rebounds, always take velocity in opposite direction as negative. This doubles the change in momentum.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem is based on the impulse-momentum theorem. The force exerted by the wall on the body causes a change in the body's momentum. The average force is equal to the rate of change of momentum. By Newton's third law, the force exerted by the body on the wall is equal in magnitude and opposite in direction.
Step 2: Key Formula or Approach:
The impulse-momentum theorem states that the average force (\(F_{avg}\)) acting on an object is equal to the change in its momentum (\(\Delta p\)) divided by the time interval (\(\Delta t\)) over which the force acts.
\[ F_{avg} = \frac{\Delta p}{\Delta t} = \frac{m v_f - m v_i}{\Delta t} \] where \(v_f\) is the final velocity and \(v_i\) is the initial velocity. Remember that velocity is a vector, so direction is important.
Step 3: Detailed Explanation:
We are given:
- Mass of the body, \( m = 5 \) kg.
- Speed before and after collision = 50 ms\(^{-1}\).
- Time of contact, \( \Delta t = \frac{1}{20} \) s.
Let's define the direction towards the wall as positive.
- Initial velocity, \( v_i = +50 \) ms\(^{-1}\).
The body rebounds, so its final velocity is in the opposite direction.
- Final velocity, \( v_f = -50 \) ms\(^{-1}\).
Now, calculate the change in momentum of the body:
\[ \Delta p = m(v_f - v_i) = 5 \text{ kg} \times (-50 \text{ ms}^{-1} - 50 \text{ ms}^{-1}) \] \[ \Delta p = 5 \text{ kg} \times (-100 \text{ ms}^{-1}) = -500 \text{ kg} \cdot \text{ms}^{-1} \] Next, calculate the average force exerted by the wall on the body:
\[ F_{\text{on body}} = \frac{\Delta p}{\Delta t} = \frac{-500 \text{ kg} \cdot \text{ms}^{-1}}{\frac{1}{20} \text{ s}} \] \[ F_{\text{on body}} = -500 \times 20 \text{ N} = -10000 \text{ N} = -1 \times 10^4 \text{ N} \] The negative sign indicates that the force exerted by the wall on the body is directed away from the wall.
The question asks for the force exerted *on the wall* (by the body). According to Newton's third law, this force is equal in magnitude and opposite in direction to the force on the body.
\[ F_{\text{on wall}} = -F_{\text{on body}} = -(-1 \times 10^4 \text{ N}) = +1 \times 10^4 \text{ N} \] The magnitude of the force is \( 1 \times 10^4 \) N.
Step 4: Final Answer:
The force exerted on the wall is \( 1 \times 10^4 \) N. This corresponds to option (D).