AP Physics 1: Unit 3 Practice Test — Circular Motion & Gravitation (2024)

The Free body diagram of the object at any general point of the circle is,

$T-mg \cosθ = \dfrac{mv^2}{r} $

Centripetal force is a function of the angle $θ$ (component of weight).
So, Option A is true.

And, Centripetal Force $= \dfrac{mv^2}{r}$
Thus, Centripetal Force changes whenever $v$ changes
So, Option D is true

For Marble,
Tension $= T = \dfrac{m_1 v^2}{r}$

For Block,
Tension $= T = m_2 g$

Thus, $\dfrac{m_1 v^2}{r} = m_2 g → v^2 $ $ = \dfrac{m_2 gr}{m_1}$

Or $v = \sqrt{\dfrac{m_2 gr}{m_1}}$

$g = \dfrac{GM}{R_e^2}$

While,
$g_h = \dfrac{GM}{(R_e+2R_e )^2} $ $ = \dfrac{GM}{9R_e^2}$

$\dfrac{g_h}{g} = \dfrac{GM}{9R_e^2}÷\dfrac{GM}{R_e^2} $ $ = \dfrac{1}{9}$

Thus, $\dfrac{g_h}{g} = 11.1 \% $

Force of gravity on $10\ m$ is equal to the vector sum of the forces from the left and bottom by $m$ and $m$

Force from left $F_l = \dfrac{10Gm^2}{y^2}$

Force from bottom $F_b = \dfrac{10Gm^2}{y^2}$

Net force = $\sqrt{F_l^2+F_b^2} = \dfrac{10\sqrt{2} Gm^2}{y^2}$
(along the diagonal)

An object feels its own weight due to the normal force acting on it.

Since inside a satellite all the objects are in free fall towards the Earth (each is moving with the same acceleration).

Hence, there are no normal forces acting on an object in a satellite.

Thus, the apparent weight is 0.

Gravitational Force = Centripetal Force

$\dfrac{Gm^2}{4a^2} = \dfrac{mv^2}{a}$

$v = \dfrac{1}{2} \sqrt{\dfrac{Gm}{a}}$

Time period $T = \dfrac{2πa}{v} $ $ = 2πa÷\dfrac{1}{2} \sqrt{\dfrac{Gm}{a}}$

$T = 4πa\sqrt{\dfrac{a}{Gm}}$

Thus, in time $δt$ the system is most likely to be in the same state
$\dfrac{δt}{(T/2)} = \dfrac{δt}{2πa} \sqrt{\dfrac{Gm}{a}}$ times.

Mass of the book $= \dfrac{28}{10} = 2.8\ kg$

(Using $g_e = 10\ m/s^2$)

Gravitational acceleration on the planet
$= \dfrac{GM_p}{R_p^2} = \dfrac{3GM_e}{4R_e^2} = 0.75\ g$

Weight on the planet
$= 0.75\ mg $ $ = 0.75×10×2.8 $ $ = 21\ N$

$GPE = \dfrac{GMm}{r}$ for $ \, r ≥ R$ ($R$ is the radius of the planet)

(We are neglecting – sign as we need only the magnitude.)

Option A → Energy should decrease with increase in distance.
Option B → Energy should decrease to 0.
Option C → Correct plot.
Option D → Energy should decrease with increase in distance.

The free body diagram for the car is,

Sum of vertical forces
$= mg-N \cos⁡α $ $ = 0 → mg $ $ = N \cos⁡\ α$

Sum of horizontal forces
$= N \sin\ α = \dfrac{mu^2}{R}$

Thus, $\tan\ ⁡α = \dfrac{u^2}{Rg}$

For lowest possible speed,
The free body diagram would be as follows,

(Friction would prevent the car to slide inwards due to less speed)

Sum of vertical forces
$= N \cos α+F_k \sin ⁡α-mg = 0$
$= 0 → N \cos α+F_k \sin α $ $ = mg$

Sum of horizontal forces
$= N \sin ⁡α-F_k \cos ⁡α $ $ = \dfrac{mu^2}{R}$

On substituting
$F_k = μ_k N$ in the above two equations,

$N(\cos α+μ_k \sin ⁡α)$ $ = mg \ \text{ and } \ N(\sin ⁡α-μ_k \cos α ) $ $ = \dfrac{mu^2}{R}$

Solving for $u$,
$\dfrac {\sin α - μ_k \cos ⁡α}{\cos ⁡α + μ_k \sin α } = \dfrac{u^2}{Rg}$

Thus, $u_{min} $ $ = \sqrt{Rg (\dfrac{\tan⁡ α - μ_k}{1 + μ_k \tan⁡ α })} $