# AP Physics C Mechanics: Unit 2 Practice Test — Newton’s Laws of Motion

Below is our AP Physics C unit 2 practice test. It covers Newton’s laws of motion and their application to solving problems with the help of free-body diagrams. Mathematical difficulty needed to solve the equations comprise of basic algebra, trigonometry, and a bit of calculus (Integration, differentiation and solving simple differential equations).

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 Question 1

### What is the acceleration of the object, in $m/s^2$, if its mass is $4 \ kg$?

 A $\dfrac{3}{4} (\skew{2.3}\hat{i} + \skew{4.5}\hat{j} )$ B $\dfrac{1}{6} (12 \skew{2.3}\hat{i} +10 \skew{4.5}\hat{j} )$ C $\dfrac{1}{4} (11 \skew{2.3}\hat{i} +7 \skew{4.5}\hat{j} )$ D $\dfrac{1}{4} (6 \skew{2.3}\hat{i} +9 \skew{4.5}\hat{j} )$ E $\dfrac{1}{4} (7 \skew{2.3}\hat{i} +11 \skew{4.5}\hat{j} )$
Question 1 Explanation:
Net force acting on the object

$=(2 \skew{2}\hat{i} +7 \skew{5}\hat{j} )+(5\skew{2}\hat{i}+4 \skew{5}\hat{j} )$

$=(7 \skew{2}\hat{i} +11 \skew{5}\hat{j} )$

Acceleration of the object

$=\dfrac{1}{mass} × Force$

$=\dfrac{1}{4} (7 \skew{2}\hat{i} +11 \skew{5}\hat{j} )$
 Question 2

### The system is released from rest. The acceleration of the $2 \ kg$ block is,

 A $2.5 \ m/s^2$ B $2.0 \ m/s^2$ C $3.6 \ m/s^2$ D $4.3 \ m/s^2$ E $5.4 \ m/s^2$
Question 2 Explanation:
The free body diagram of the two blocks is,

$T-2g=2a$ … (1)

$3g-T=3a$ … (2)

$g = 5a → a = \dfrac{g}{5} =2 \ m/s^2$
 Question 3

### The marble is set into motion at a uniform speed $u$. Which of the below forces keeps the marble moving at a uniform speed?

 A Friction between marble and the ring B Weight of the marble C Weight of the ring D Normal force due to the ring E None of the above
Question 3 Explanation:
Check each option,

Option A → Incorrect, as friction would slow down the marble in some time

Option B → Incorrect, as the weight of the marble is perpendicular to the plane of the velocity

Option C → Incorrect, as the weight of the ring cannot affect the motion of the marble directly

Option D → Correct, as the normal force is always perpendicular to the velocity of the marble and is in the same plane

Option E → Incorrect, as D is correct
 Question 4

### A block of mass $2 \ kg$ is on an inclined plane. The coefficient of static friction between the block and the plane is $0.4$. How much force would be needed to set the block in motion (down the incline) if the angle of the inclined plane is $12°$?

 A $3.67 \ N$ B $7.82 \ N$ C $5.22 \ N$ D $4.15 \ N$ E $2.33 \ N$
Question 4 Explanation:
In equilibrium, the free body diagram of the block is as follows:

For equilibrium,

$N = 2g \cos 12°$ and $f_r = 2g \ \sin⁡ 12°$

Using $g = 10 \ m/s^2,$

$N = 19.56 \ N$ and $f_r = 4.15 \ N$

In static condition, $f_r ≤ μN$ (which is true for the system)

Extra force needed to set the block in motion (down the incline)

$=μN-f_r$

$=0.4 × 19.56 - 4.15$

$=7.82 - 4.15$

$=3.67 \ N$
 Question 5

### If the coefficient of friction for the right block is $0.33$, then what is the value of the coefficient of friction $μ$ for the left block? (Assume the system remains just in equilibrium and any small force can set the system in motion.)

 A $0.250$ B $0.333$ C $0.223$ D $0.252$ E $0.283$
Question 5 Explanation:
The free body diagram for the two-block system is given below:

For the right block: $25 = T + f_1$ and $N = 5g =50$

Combining the two equations:

$25 = T +0.33 × 50$

$T = 8.5 \ N$

For the left block: $T = f_2$ and $N = 3g = 30$

Combining the two equations:

$8.5 = 30μ$

$μ = 0.283$
 Question 6

### The coefficient of static friction between the two blocks is $0.25$, and the coefficient of kinetic friction is $0.18$. If a force of $40 \ N$ acts on the lower block, what will be the acceleration of the upper block? (Assume there is no friction between the lower block and the floor)

 A $2.25 \ m/s^2$ B $2.85 \ m/s^2$ C $1.80 \ m/s^2$ D $1.55 \ m/s^2$ E $1.20 \ m/s^2$
Question 6 Explanation:
Assuming the two blocks move together with the same acceleration a,

$a = \dfrac{F}{m_{down} + m_{up}}$

$=\dfrac{40}{14}$

$=2.85 \ m/s^2$

The free body diagram for the upper block is:

Thus, for the block to move with an acceleration of $2.85 \ m/s^2$ the frictional force must be at least $4 × 2.85 = 11.4 \ N$

Maximum value of static friction $= 0.25 × 40 = 10 N < 11.4 \ N$

Thus, the upper block slides with acceleration lesser than the lower block

$a = \dfrac{μ_k \ N}{m}$

$= \dfrac{0.18×40}{4}$

$= 1.8 \ m/s^2$
 Question 7

### A particle of mass $m$ is moving through a horizontally oriented medium. The particle experiences a drag force of $γv$ in a direction opposite to the velocity. If there is no other force acting on the particle, then which of the below equations represents the motion of the particle?

 A $m \dfrac{dv}{dt} - γv = 0$ B $m \dfrac{dv}{dt} + γv = 0$ C $m \dfrac{dv}{dt} - γv = ma$ D $m \dfrac{dv}{dt} + 2γv = 0$ E None of the above
Question 7 Explanation:
$F = m \dfrac{dv}{dt}$

The only force acting on the particle is the drag force of the form $-γv$

Thus, $m \dfrac{dv}{dt} = -γv$

Or $m \dfrac{dv}{dt} + γv = 0$
 Question 8

### A stone of mass $m$ tied to a massless string of length $r$ is set into motion. If the stone moves in a vertical circle, what is the minimum velocity that the stone must have when it is at the topmost point of the circle?

 A $\sqrt{5gr}$ B $\sqrt{4gr}$ C $\sqrt{3gr}$ D $\sqrt{2gr}$ E $\sqrt{gr}$
Question 8 Explanation:
For minimum velocity, Tension force $= 0 \ N$

At the topmost point, the centripetal force is provided by the gravitational force of the stone

$mg = \dfrac{mv^2}{r} → v = \sqrt{rg}$
 Question 9

### What is the average force acting on the car from $t = 1$ to $t = 4$?

 A $2m(\skew{2.3}\hat{i} + \skew{4.5}\hat{j} )$ B $5m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j} )$ C $4m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j})$ D $m(2\skew{2.3}\hat{i} - 3 \skew{4.5}\hat{j} )$ E $3m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j})$
Question 9 Explanation:
$\overrightarrow{v}(t = 1)=5 \skew{2}\hat{i} + 9 \skew{5}\hat{j}$ and $\overrightarrow{v}(t=4)=20 \skew{2}\hat{i} -6 \skew{5}\hat{j}$

Average acceleration

$=\dfrac{\overrightarrow{v}(t=4) - \overrightarrow{v}(t=1)}{4 - 1}$

$=\dfrac{15 \skew{2}\hat{i} - 15 \skew{5}\hat{j} }{3}$

$=5 \skew{2}\hat{i}-5 \skew{5}\hat{j}$

Average force $= m × \overrightarrow{a}=5m( \skew{2}\hat{i} - \skew{5}\hat{j} )$
 Question 10

### The strings and the pulley are massless and the frictional force between the blocks and the surface is zero. When the system starts from rest, the tension force between the $4 \ kg$ block and the $2 \ kg$ block is,

 A $20 \ N$ B $12 \ N$ C $15 \ N$ D $10 \ N$ E $8 \ N$
Question 10 Explanation:
The entire system moves with the same magnitude of acceleration

For the block moving vertically: $2g - T = 2a$

For the two-block system moving horizontally: $T = 6a$

Using the two equations, $2g - 6a = 2a → a = \dfrac{g}{4}= 2.5 \ m/s^2$

For the 4 kg block, the equation of motion is:

$T_1 = 4a → T_1 = 10 \ N$
 Question 11

### What is the acceleration of the $2 \ kg$ block kept on the table?

 A $2.5 \ m/s^2$ B $2.0 \ m/s^2$ C $3.3 \ m/s^2$ D $4.2 \ m/s^2$ E $1.8 \ m/s^2$
Question 11 Explanation:
From question 10,

$a = \dfrac{g}{4}$

Thus, the acceleration of the system of blocks is equal to

$2.5 \ m/s^2$
 Question 12

### At $t=t_{break}$, the string between the $4 \ kg$ block and the $2 \ kg$ block breaks. What will be the acceleration of the $2 \ kg$ block now?

 A $3 \ m/s^2$ B $4 \ m/s^2$ C $5 \ m/s^2$ D $6 \ m/s^2$ E None of the above
Question 12 Explanation:
After $t=t_{break}$,

For the block moving vertically: $2g-T=2a$

For the one-block system moving horizontally: $T=2a$

Using the two equations, $2g-2a=2a→a=\dfrac{g}{2}=5 \ m/s^2$
 Question 13

### As a result, the block moves upwards and travels a distance $h$ before it starts sliding down. The coefficient of friction between the block and the inclined plane is $0.15$, and the angle of inclination is $25°$. What is the value of $h$?

 A $34.56 \ m$ B $30.60 \ m$ C $32.04 \ m$ D $29.03 \ m$ E $22.16 \ m$
Question 13 Explanation:
The free body diagram of the block is:

The equations of motion are:

$N=g \ \cos⁡ 25°$ and $g \ \sin⁡ 25° +f_r=a$

Using $f_r=0.15×N$ in the above equations gives,

$a=g \sin⁡ 25° +0.15g \ \cos⁡ 25°$

$a=g(\sin⁡ 25° +0.15 \ \cos⁡ 25° )$

$a=5.58 \ m/s^2$

Using $a=5.58 \ m/s^2, u=18 \ m/s$ and $v=0 \ m/s$ in $v^2=u^2-2ah$ gives,

$h=\dfrac{u^2}{2a} = \dfrac{18^2}{2×5.58}$

$h=29.03 \ m$
 Question 14

### The object moves in a circle of radius 3 cm. Find the speed of the object when $t = 6 \ s$. (Note: The angles are in radians)

 A $0.980 \ m/s$ B $0.060 \ m/s$ C $0.802 \ m/s$ D $2.451 \ m/s$ E $0.353 \ m/s$
Question 14 Explanation:
$\overrightarrow{f}_{cent} (t=6)=10[\cos \ ⁡48 \skew{2}\hat{i} + \sin \ ⁡48 \skew{5}\hat{j} ]$

Thus, the magnitude of the centripetal force is

$\sqrt{10^2 (\sin^2⁡ \ 48 + \cos^2 \ ⁡48)} =10 \ N$

Since, $F = \dfrac{mv^2}{r} → v = \sqrt{ \dfrac{Fr}{m}}$

$v = \sqrt{10 × \dfrac{0.03}{2.4}}=0.353 \ m/s$
 Question 15

### Where $A$ is a positive constant and $v(t)$ is the time-dependent velocity of the object. Find an expression for $v(t$) subject to the condition that $v(0)=1 \ m/s$.

 A $\left(1+\dfrac{2At}{m}\right)^{-1/2}$ B $\left(1+\dfrac{2At}{m}\right)^{-2}$ C $\left(1+\dfrac{2At}{m}\right)^{-1}$ D $\left(1+\dfrac{4At}{m}\right)^{1/2}$ E $\left(1+\dfrac{4At}{m}\right)^{2}$
Question 15 Explanation:
$m \dfrac{dv}{dt} = -Av^3$

Using separation of variables,

$\dfrac{dv}{v^3} = -\dfrac{A}{m} \ dt$

Integrating both sides gives,

$\int \dfrac{dv}{v^3} =-\int \dfrac{A}{m} \ dt$

$\dfrac{v^{-2}}{-2} = - \dfrac{A}{m} t-c$ (Where $c$ is some constant)

$v^{-2} = \dfrac{2A}{m} t+c'$ (Where $c'$ is some other constant)

Using $v(t=0)=1 \ m/s$ gives,

$c'=1$

Thus, $v(t) = \left(1 + \dfrac{2At}{m}\right)^{-1/2}$
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