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 |

### Two independent sources of force act on an object. The vector form of these forces is given below:

### $\overrightarrow{F}_I = 2 \skew{2.5}\hat{i} + 7 \skew{4}\hat{j} \ N$ and $\overrightarrow{F}_{II} = 5 \skew{2.5}\hat{i} + 4 \skew{4}\hat{j} \ N$

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

$ \dfrac{3}{4} (\skew{2.3}\hat{i} + \skew{4.5}\hat{j} ) $ | |

$ \dfrac{1}{6} (12 \skew{2.3}\hat{i} +10 \skew{4.5}\hat{j} )$ | |

$ \dfrac{1}{4} (11 \skew{2.3}\hat{i} +7 \skew{4.5}\hat{j} )$ | |

$ \dfrac{1}{4} (6 \skew{2.3}\hat{i} +9 \skew{4.5}\hat{j} ) $ | |

$ \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} ) $

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

$=(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 |

### Look at the below Atwood Machine with a frictionless pulley. Assume the string and the pulley are of negligible mass.

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

$2.5 \ m/s^2$ | |

$2.0 \ m/s^2$ | |

$3.6 \ m/s^2$ | |

$4.3 \ m/s^2$ | |

$5.4 \ m/s^2$ |

Question 2 Explanation:

The free body diagram of the two blocks is,

$T-2g=2a$ … (1)

$3g-T=3a$ … (2)

Adding (1) and (2),

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

$T-2g=2a$ … (1)

$3g-T=3a$ … (2)

Adding (1) and (2),

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

Question 3 |

### A frictionless circular shaped ring has a small marble of mass $m$ fixed on it, as shown below.

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

Friction between marble and the ring | |

Weight of the marble | |

Weight of the ring | |

Normal force due to the ring | |

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

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 $12°$?

$3.67 \ N$ | |

$7.82 \ N$ | |

$5.22 \ N$ | |

$4.15 \ N$ | |

$2.33 \ N$ |

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)

$=μN-f_r$

$=0.4 × 19.56 - 4.15$

$=7.82 - 4.15$

$=3.67 \ N$

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 |

### Two blocks are connected by a massless string of length $l$. A force of $25 \ N$ acts on the right block, as shown below.

### 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.)

$0.250$ | |

$0.333$ | |

$0.223$ | |

$0.252$ | |

$0.283$ |

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:

$8.5 = 30μ$

$μ = 0.283$

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 |

### Two blocks are kept on each other, as shown below.

### 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)

$2.25 \ m/s^2$ | |

$2.85 \ m/s^2$ | |

$1.80 \ m/s^2$ | |

$1.55 \ m/s^2$ | |

$1.20 \ m/s^2$ |

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 $

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

$=\dfrac{40}{14}$

$=2.85 \ m/s^2$

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?

$m \dfrac{dv}{dt} - γv = 0$ | |

$m \dfrac{dv}{dt} + γv = 0$ | |

$m \dfrac{dv}{dt} - γv = ma$ | |

$m \dfrac{dv}{dt} + 2γv = 0$ | |

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 $m \dfrac{dv}{dt} + γv = 0$

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?

$\sqrt{5gr}$ | |

$\sqrt{4gr}$ | |

$\sqrt{3gr}$ | |

$\sqrt{2gr}$ | |

$\sqrt{gr}$ |

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} $

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 |

### The velocity vector of a car of mass m is given by the equation:

### $\overrightarrow{v}(t) = 5t \skew{2.5}\hat{i} + (10 -t^2) \skew{4}\hat{j}$

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

$2m(\skew{2.3}\hat{i} + \skew{4.5}\hat{j} )$ | |

$5m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j} )$ | |

$4m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j})$ | |

$m(2\skew{2.3}\hat{i} - 3 \skew{4.5}\hat{j} )$ | |

$3m(\skew{2.3}\hat{i} - \skew{4.5}\hat{j})$ |

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

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 |

**Questions 10, 11 and 12 are based on the below information:**

### Look at the below system of three blocks.

### 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,

$20 \ N$ | |

$12 \ N$ | |

$15 \ N$ | |

$10 \ N$ | |

$8 \ N$ |

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 $

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 |

### Look at the below system of three blocks.

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

$2.5 \ m/s^2$ | |

$2.0 \ m/s^2$ | |

$3.3 \ m/s^2$ | |

$4.2 \ m/s^2$ | |

$1.8 \ m/s^2$ |

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 $

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

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

$2.5 \ m/s^2 $

Question 12 |

### Look at the below system of three blocks.

### 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?

$3 \ m/s^2$ | |

$4 \ m/s^2$ | |

$5 \ m/s^2$ | |

$6 \ m/s^2$ | |

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, $2g-2a=2a→a=\dfrac{g}{2}=5 \ m/s^2$

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 |

### A block at the bottom of an inclined plane is given a sharp push at $t = 0 \ s$.

### 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 $h$?

$34.56 \ m$ | |

$30.60 \ m$ | |

$32.04 \ m$ | |

$29.03 \ m$ | |

$22.16 \ m$ |

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,

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

$h=29.03 \ m$

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 |

### A time-dependent centripetal force acts on an object of mass $2.4 \ kg$. The expression for the centripetal force is given as,

### $\overrightarrow{f}_{cent} (t)=10[\cos 8t \skew{2.5}\hat{i} + \sin 8t \skew{4}\hat{j} ] N$

### 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)

$0.980 \ m/s$ | |

$0.060 \ m/s$ | |

$0.802 \ m/s$ | |

$2.451 \ m/s$ | |

$0.353 \ m/s$ |

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 $

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 |

### The differential equation of an object moving through a viscous medium is given as:

### $m \dfrac{dv}{dt} = -Av^3$

### 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 $v(0)=1 \ m/s$.

$\left(1+\dfrac{2At}{m}\right)^{-1/2}$ | |

$\left(1+\dfrac{2At}{m}\right)^{-2}$ | |

$\left(1+\dfrac{2At}{m}\right)^{-1}$ | |

$\left(1+\dfrac{4At}{m}\right)^{1/2}$ | |

$\left(1+\dfrac{4At}{m}\right)^{2}$ |

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

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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