3. A ball is rolling on a frictionless surface at $4 \, \text{m/s}$. A constant force of $2 \, \text{N}$ is applied in the direction opposite to its motion for $3 \, \text{s}$. What happens to the ball after the force is applied?
Key Concept: Force, Multiple Effects
c) The ball slows down and may reverse direction depending on its mass.
[Solution Description]
The question combines concepts of force, deceleration, and change in motion. On a frictionless surface, the only force acting is the applied $2 \, \text{N}$. Using Newton’s second law ($F = ma$), we find the deceleration ($a = F/m$, assuming mass $m$). The change in velocity ($v = u + at$) will show how the ball’s speed and direction are affected over the $3 \, \text{s}$ interval.
Step-by-step calculations:
1. Deceleration: $a = F/m = 2 \, \text{N} / m$.
2. Final velocity: $v = 4 \, \text{m/s} - (2/m \times 3) = 4 - 6/m \, \text{m/s}$.
3. If $6/m > 4$, the ball stops and reverses direction; else, it slows down.
Your Answer is correct.
c) The ball slows down and may reverse direction depending on its mass.
[Solution Description]
The question combines concepts of force, deceleration, and change in motion. On a frictionless surface, the only force acting is the applied $2 \, \text{N}$. Using Newton’s second law ($F = ma$), we find the deceleration ($a = F/m$, assuming mass $m$). The change in velocity ($v = u + at$) will show how the ball’s speed and direction are affected over the $3 \, \text{s}$ interval.
Step-by-step calculations:
1. Deceleration: $a = F/m = 2 \, \text{N} / m$.
2. Final velocity: $v = 4 \, \text{m/s} - (2/m \times 3) = 4 - 6/m \, \text{m/s}$.
3. If $6/m > 4$, the ball stops and reverses direction; else, it slows down.