AP Physics 1 - Advanced Concepts & Hard Problems

Front

Why can't you use the kinematic equation $v_f = v_i + a \Delta t$ for the entire flight of a ball thrown in the air if air resistance is significant?

Back

This equation assumes **constant acceleration**. With air resistance, acceleration is **not constant**; it depends on velocity (often proportional to $v$ or $v^2$). As the ball rises, slows down, and falls, the direction and magnitude of the drag force change, causing acceleration to vary continuously throughout the motion.

Front

In a collision with a wall, how can momentum be conserved for a ball if the ball reverses direction and the wall remains stationary?

Back

Momentum is a vector quantity. While the ball's momentum changes from $+p$ to $-p$, the **Earth-Wall system** absorbs the impulse. Since the wall is attached to the Earth (mass $M \gg m_{ball}$), the wall's velocity change is imperceptibly small ($\Delta v_{wall} \approx 2m_{ball}v_{ball}/M$), satisfying conservation: $\Delta \vec{p}_{ball} + \Delta \vec{p}_{wall} = 0$.

Front

Compare the tension force at the top ($T_{top}$) versus the bottom ($T_{bottom}$) of a vertical circle swung at constant speed.

Back

At the bottom, Tension must support gravity and provide centripetal force: $T_b = mg + \frac{mv^2}{r}$. At the top, gravity helps provide centripetal force, so Tension is reduced: $T_t = \frac{mv^2}{r} - mg$. Therefore, **$T_{bottom}$ is always greater than $T_{top}$** by $2mg$.

Front

How does the period of a simple pendulum change when the amplitude is small vs. large?

Back

For **small angles** ($<15^\circ$), the period is isochronous and independent of amplitude ($T \approx 2\pi\sqrt{L/g}$). For **large amplitudes**, the restoring force is no longer proportional to the displacement (simple harmonic motion fails), and the period **increases** slightly as amplitude increases.

Front

If a block slides down a frictionless incline, does the Normal Force do work?

Back

**No.** The work done by a force is $W = F d \cos(\theta)$. The Normal Force is always perpendicular to the surface, while the displacement is parallel to the surface. Since the angle between them is $90^\circ$ and $\cos(90^\circ) = 0$, the work done by the Normal Force is zero.