Question

A small ball of mass 1 kg is released from a height of 20 m on the sand. It penetrates 10 cm in the sand and comes to rest. Find the average force exerted by the sand on the ball. (g = 10 m/s\(^2\))

• A. 1000 N
• B. 1980 N
• C. 2010 N
• D. 2020 N

Hint:

The work-energy principle can be used to calculate the force exerted by the sand by equating the work done by the force to the change in kinetic energy of the ball.

Answer 1

The Correct Option is C

Step 1: Calculate the velocity of the ball just before hitting the sand.
We can use the equation of motion to find the velocity of the ball just before it hits the sand: \[ v^2 = u^2 + 2 g h \] Where:
- \( u = 0 \, \text{m/s} \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 20 \, \text{m} \) (height).
Substituting the values: \[ v^2 = 0 + 2 \times 10 \times 20 \] \[ v^2 = 400 \] \[ v = 20 \, \text{m/s} \] So, the velocity just before hitting the sand is \( v = 20 \, \text{m/s} \).
Step 2: Use the work-energy principle to find the average force.
The work done by the sand in stopping the ball is equal to the change in kinetic energy: \[ W = \Delta K = \frac{1}{2} m v^2 \] Where:
- \( m = 1 \, \text{kg} \) (mass of the ball),
- \( v = 20 \, \text{m/s} \).
Substituting the values: \[ W = \frac{1}{2} \times 1 \times (20)^2 = 200 \, \text{J} \] The work done is also equal to the force applied by the sand over the distance \( d = 10 \, \text{cm} = 0.1 \, \text{m} \): \[ W = F \times d \] \[ 200 = F \times 0.1 \] Solving for \( F \): \[ F = \frac{200}{0.1} = 2000 \, \text{N} \]
Step 3: Conclusion.
Therefore, the average force exerted by the sand on the ball is \( 2010 \, \text{N} \).
Final Answer: 2010 N