Question

A soccer ball of mass 250 g is moving horizontally to the left with a speed of 22 m/s. This ball is kicked towards right with a velocity 30 m/s at an angle 53° with the horizontal in upward direction. Assuming that it took 0.01 s for the collision to take place, the average force acting is:

• A. 1000 N
• B. 986 N
• C. 1166 N
• D. 2000 N

Hint:

For problems involving change in velocity and impulse, break the velocity into components and calculate the resultant change in velocity. Then calculate the force using impulse-momentum theorem.

Answer 1

The Correct Option is C

To determine the average force acting on the soccer ball during the collision, we need to calculate the change in momentum and then use the impulse
-momentum theorem. Given: 
- Mass of the ball, \( m = 250 \, \text{g} = 0.25 \, \text{kg} \) 
- Initial velocity, \( \vec{v}_i = -22 \, \text{m/s} \) (to the left) 
- Final velocity, \( \vec{v}_f = 30 \, \text{m/s} \) at \( 53^\circ \) above the horizontal to the right 
- Time of collision, \( \Delta t = 0.01 \, \text{s} \) Step 1: Resolve the Final Velocity into Components The final velocity \( \vec{v}_f \) can be resolved into horizontal (\( v_{fx} \)) and vertical (\( v_{fy} \)) components: \[ v_{fx} = 30 \cos 53^\circ \] \[ v_{fy} = 30 \sin 53^\circ \] Using trigonometric values: \[ \cos 53^\circ \approx 0.6 \quad \text{and} \quad \sin 53^\circ \approx 0.8 \] Thus: \[ v_{fx} = 30 \times 0.6 = 18 \, \text{m/s} \] \[ v_{fy} = 30 \times 0.8 = 24 \, \text{m/s} \] Step 2: Calculate the Change in Momentum The initial momentum \( \vec{p}_i \) and final momentum \( \vec{p}_f \) are: \[ \vec{p}_i = m \vec{v}_i = 0.25 \times (-22) = -5.5 \, \text{kg} \cdot \text{m/s} \] \[ \vec{p}_f = m \vec{v}_f = 0.25 \times (18 \hat{i} + 24 \hat{j}) = 4.5 \hat{i} + 6 \hat{j} \, \text{kg} \cdot \text{m/s} \] The change in momentum \( \Delta \vec{p} \) is: \[ \Delta \vec{p} = \vec{p}_f - \vec{p}_i = (4.5 \hat{i} + 6 \hat{j}) - (-5.5 \hat{i}) = 10 \hat{i} + 6 \hat{j} \, \text{kg} \cdot \text{m/s} \] Step 3: Calculate the Magnitude of the Change in Momentum \[ |\Delta \vec{p}| = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.66 \, \text{kg} \cdot \text{m/s} \] Step 4: Calculate the Average Force Using the impulse-momentum theorem: \[ \vec{F}_{\text{avg}} = \frac{\Delta \vec{p}}{\Delta t} = \frac{11.66}{0.01} = 1166 \, \text{N} \] Final Answer: \[ \boxed{1166 \, \text{N}} \] This corresponds to option (3).