Step-by-Step Physics Study Guidance for Quizzes 1 & 2

Quiz 1 – Chapters 1 & 2

Q1. When a ball is thrown straight up with no air resistance, what is the acceleration at its highest point?

Background

Topic: Kinematics and Free Fall

This question tests your understanding of acceleration due to gravity, especially at the peak of a projectile's motion.

Key Terms and Concepts:

• Acceleration due to gravity ($g$): The constant acceleration experienced by objects in free fall near Earth's surface, directed downward.

• Highest point: The instant when the object's velocity is zero, but acceleration may not be.

Step-by-Step Guidance

1. Recall that when an object is thrown upward, gravity acts on it throughout its flight.

2. At the highest point, the velocity of the object is zero, but consider whether the acceleration is also zero or if it remains constant.

3. Think about the direction of the acceleration vector at all points during the motion, including the peak.

Try solving on your own before revealing the answer!

Q2. If a car accelerates at a uniform 4.0 m/s², how long will it take to reach a speed of 80 m/s, starting from rest?

Background

Topic: Linear Kinematics (Constant Acceleration)

This question tests your ability to use kinematic equations to relate acceleration, initial velocity, final velocity, and time.

Key Formula:

$v = v_0 + at$

• $v$ = final velocity

• $v_0$ = initial velocity (starting from rest means $v_0 = 0$)

• $a$ = acceleration

• $t$ = time

Step-by-Step Guidance

1. Identify the known values: $v_0 = 0$, $v = 80\ \mathrm{m/s}$, $a = 4.0\ \mathrm{m/s}^2$.

2. Write the kinematic equation: $v = v_0 + at$.

3. Substitute the known values into the equation.

4. Rearrange the equation to solve for $t$.

Try solving on your own before revealing the answer!

Q3. A car initially traveling at 60 m/s accelerates at a constant rate of 2.0 m/s². How much time is required for the car to reach a speed of 90 m/s?

Background

Topic: Linear Kinematics (Constant Acceleration)

This question tests your ability to use the kinematic equation for velocity when acceleration is constant.

Key Formula:

$v = v_0 + at$

• $v_0$ = initial velocity ($60\ \mathrm{m/s}$)

• $v$ = final velocity ($90\ \mathrm{m/s}$)

• $a$ = acceleration ($2.0\ \mathrm{m/s}^2$)

• $t$ = time

Step-by-Step Guidance

1. List the known values: $v_0 = 60\ \mathrm{m/s}$, $v = 90\ \mathrm{m/s}$, $a = 2.0\ \mathrm{m/s}^2$.

2. Write the equation: $v = v_0 + at$.

3. Plug in the values for $v$, $v_0$, and $a$.

4. Rearrange to solve for $t$.

Try solving on your own before revealing the answer!

Q4. A car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s² for 15 s. At the end of this time, what is its velocity?

Background

Topic: Kinematics – Multiple Motion Segments

This question tests your ability to analyze motion in stages: constant velocity followed by constant acceleration.

Key Formulas:

• For constant velocity: $v = \text{constant}$

• For constant acceleration: $v = v_0 + at$

Step-by-Step Guidance

1. First, note that during the initial 10 s, the velocity remains $15\ \mathrm{m/s}$.

2. After 10 s, the car accelerates at $2.0\ \mathrm{m/s}^2$ for $15\ \mathrm{s}$, starting from $v_0 = 15\ \mathrm{m/s}$.

3. Use $v = v_0 + at$ to find the final velocity after the acceleration phase.

4. Plug in $v_0 = 15\ \mathrm{m/s}$, $a = 2.0\ \mathrm{m/s}^2$, and $t = 15\ \mathrm{s}$.

Try solving on your own before revealing the answer!

Q5. A car starting from rest accelerates at a constant 2.0 m/s² for 10 s. It then travels with the constant speed it has achieved for another 10 s. Then it finally slows to a stop with a constant acceleration of magnitude 2.0 m/s². How far does it travel after starting?

Background

Topic: Kinematics – Piecewise Motion

This question tests your ability to break a complex motion into segments and use kinematic equations for each part.

Key Formulas:

• For constant acceleration: $d = v_0 t + \frac{1}{2} a t^2$

• For constant velocity: $d = v t$

• For deceleration: Use kinematic equations with negative acceleration.

Step-by-Step Guidance

1. Segment 1: Calculate the distance covered during the initial acceleration ($a = 2.0\ \mathrm{m/s}^2$, $t = 10\ \mathrm{s}$, $v_0 = 0$).

2. Find the velocity at the end of the first segment using $v = v_0 + at$.

3. Segment 2: Calculate the distance traveled at constant velocity for $10\ \mathrm{s}$.

4. Segment 3: Calculate the distance during deceleration to rest (use $v_0$ from segment 2, $v = 0$, $a = -2.0\ \mathrm{m/s}^2$).

5. Add the distances from all three segments to get the total distance.

Try solving on your own before revealing the answer!

Quiz 2

Q1. Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. What is the resultant vector A + B?

Background

Topic: Vector Addition

This question tests your ability to resolve vectors into components and add them to find the resultant vector's magnitude and direction.

Key Formulas:

• Component form: $A_x = A \cos \theta$, $A_y = A \sin \theta$

• Resultant components: $R_x = A_x + B_x$, $R_y = A_y + B_y$

• Magnitude: $R = \sqrt{R_x^2 + R_y^2}$

• Direction: $\theta = \arctan\left(\frac{R_y}{R_x}\right)$

Step-by-Step Guidance

1. Resolve both vectors into their $x$ and $y$ components, being careful with the angles and directions.

2. Add the corresponding components to find $R_x$ and $R_y$.

3. Use the Pythagorean theorem to find the magnitude of the resultant vector.

4. Calculate the direction of the resultant using the arctangent formula.

Try solving on your own before revealing the answer!

Q2. The figure shows four vectors A, B, C, and D, having magnitudes 12.0 m, 10.0 m, 8.0 m, and 4.0 m, respectively. The sum of these four vectors is:

Background

Topic: Vector Addition (Graphical and Analytical)

This question tests your ability to add multiple vectors, likely using components, to find the resultant's magnitude and direction.

Key Formulas:

• Sum components: $R_x = A_x + B_x + C_x + D_x$, $R_y = A_y + B_y + C_y + D_y$

• Magnitude: $R = \sqrt{R_x^2 + R_y^2}$

• Direction: $\theta = \arctan\left(\frac{R_y}{R_x}\right)$

Step-by-Step Guidance

1. Write out the $x$ and $y$ components for each vector (you may need the figure for directions).

2. Add all $x$ components to get $R_x$ and all $y$ components to get $R_y$.

3. Calculate the magnitude and direction using the formulas above.

Try solving on your own before revealing the answer!

Q3. A ball is thrown with an initial velocity of 20 m/s at an angle of 60° above the horizontal. If we can neglect air resistance, what is the horizontal component of its instantaneous velocity at the exact top of its trajectory?

Background

Topic: Projectile Motion

This question tests your understanding of how velocity components behave at the peak of a projectile's path.

Key Formulas:

• Horizontal component: $v_x = v_0 \cos \theta$

• At the top, the vertical component is zero, but the horizontal component remains unchanged (if no air resistance).

Step-by-Step Guidance

1. Identify the initial velocity $v_0 = 20\ \mathrm{m/s}$ and launch angle $\theta = 60^\circ$.

2. Recall that the horizontal velocity remains constant throughout the flight (no air resistance).

3. Calculate $v_x$ using $v_x = v_0 \cos \theta$.

Try solving on your own before revealing the answer!

Q4. A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff if air resistance is negligibly small?

Background

Topic: Projectile Motion (Horizontal Launch)

This question tests your ability to analyze vertical motion under gravity, independent of horizontal motion.

Key Formula:

$y = v_{0y} t + \frac{1}{2} g t^2$

• For horizontal launch, $v_{0y} = 0$

• $g = 9.8\ \mathrm{m/s}^2$ (downward acceleration due to gravity)

Step-by-Step Guidance

1. Recognize that the initial vertical velocity is zero ($v_{0y} = 0$).

2. Use the formula $y = \frac{1}{2} g t^2$ to find the vertical distance fallen.

3. Plug in $g = 9.8\ \mathrm{m/s}^2$ and $t = 4.3\ \mathrm{s}$.

Try solving on your own before revealing the answer!

Q5. A heavy rock is shot upward from the edge of a vertical cliff with an initial velocity of 15 m/s directed at 25° from the vertical. How high is the cliff?

Background

Topic: Projectile Motion (Non-Horizontal Launch)

This question tests your ability to analyze projectile motion launched at an angle from a height.

Key Formulas:

• Vertical motion: $y = v_{0y} t - \frac{1}{2} g t^2$

• Find $v_{0y}$ using $v_{0y} = v_0 \cos \theta$ (since angle is from vertical)

• Time of flight can be found by setting $y = -h$ (where $h$ is the height of the cliff)

Step-by-Step Guidance

1. Resolve the initial velocity into vertical and horizontal components (be careful with the angle from vertical).

2. Set up the equation for vertical displacement, considering the rock lands at the base of the cliff.

3. Use kinematic equations to relate time of flight and height.

4. Set up the equation to solve for the height of the cliff.

Try solving on your own before revealing the answer!

Q6. A boy throws a rock with an initial velocity of 3.13 m/s at 30.0° above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory if air resistance is negligibly small and g = 9.80 m/s²?

Background

Topic: Projectile Motion – Time to Maximum Height

This question tests your ability to find the time to reach the peak of a projectile's path.

Key Formula:

$t = \frac{v_{0y}}{g}$

• $v_{0y} = v_0 \sin \theta$

Step-by-Step Guidance

1. Calculate the initial vertical velocity: $v_{0y} = v_0 \sin \theta$.

2. Use the formula $t = \frac{v_{0y}}{g}$ to find the time to maximum height.

3. Plug in the values for $v_0$, $\theta$, and $g$.

Try solving on your own before revealing the answer!

Q7. An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. How long does the athlete stay in the air, assuming no significant air resistance?

Background

Topic: Projectile Motion – Time of Flight

This question tests your ability to calculate the total time a projectile spends in the air.

Key Formula:

$t_{\text{total}} = \frac{2 v_0 \sin \theta}{g}$

• $v_0$ = initial speed

• $\theta$ = launch angle

• $g$ = acceleration due to gravity

Step-by-Step Guidance

1. Calculate the initial vertical velocity: $v_{0y} = v_0 \sin \theta$.

2. Use the formula for total time of flight: $t_{\text{total}} = \frac{2 v_{0y}}{g}$.

3. Plug in the values for $v_0$, $\theta$, and $g$.

Try solving on your own before revealing the answer!

Q8. For general projectile motion with no air resistance, the vertical component of a projectile's acceleration:

Background

Topic: Projectile Motion – Acceleration Components

This question tests your understanding of how acceleration behaves in projectile motion.

Key Concept:

• The vertical acceleration is due to gravity and remains constant (unless otherwise specified).

Step-by-Step Guidance

1. Recall that gravity acts downward at all times during projectile motion.

2. Consider whether the vertical acceleration changes or remains constant throughout the flight.

Try solving on your own before revealing the answer!

Q9. In order to get an object moving, you must push harder on it than it pushes back on you. True or False?

Background

Topic: Newton's Third Law of Motion

This question tests your understanding of action-reaction force pairs.

Key Concept:

• Newton's Third Law: For every action, there is an equal and opposite reaction.

Step-by-Step Guidance

1. Think about what Newton's Third Law says about the forces two objects exert on each other.

2. Consider whether it is possible for one object to exert a greater force than the other in the interaction.

Try solving on your own before revealing the answer!

Q10. A truck is using a hook to tow a car whose mass is one quarter that of the truck. If the force exerted by the truck on the car is 6000 N, then the force exerted by the car on the truck is:

Background

Topic: Newton's Third Law of Motion

This question tests your understanding of action-reaction pairs, regardless of the masses involved.

Key Concept:

• Newton's Third Law: The force exerted by the car on the truck is equal in magnitude and opposite in direction to the force exerted by the truck on the car.

Step-by-Step Guidance

1. Identify the action force (truck on car) and the reaction force (car on truck).

2. Recall that the magnitudes of these forces are always equal, regardless of the masses.

3. Use the given value to determine the magnitude of the reaction force.

Try solving on your own before revealing the answer!

Q11. A 75-N box rests on a perfectly smooth horizontal surface. The minimum force needed to start the box moving is:

Background

Topic: Friction and Newton's Laws

This question tests your understanding of frictional forces, especially on a frictionless surface.

Key Concept:

• On a perfectly smooth (frictionless) surface, the force of friction is zero.

Step-by-Step Guidance

1. Recall that static friction opposes the start of motion, but only if friction is present.

2. Consider what happens when the surface is perfectly smooth (no friction).

3. Determine what force is required to overcome friction in this scenario.

Try solving on your own before revealing the answer!

Q12. A woman is straining to lift a large crate, but without success because it is too heavy. We denote the forces on the crate as follows: P = magnitude of the upward force exerted by the person, C = magnitude of the vertical contact force from the floor, W = weight of the crate. How are these forces related while the person is unsuccessfully trying to lift the crate?

Background

Topic: Forces in Equilibrium

This question tests your understanding of force balance when an object remains at rest.

Key Concept:

• When the crate does not move, the sum of upward forces equals the downward force (weight).

Step-by-Step Guidance

1. Draw a free-body diagram showing all vertical forces acting on the crate.

2. Set up the equilibrium condition: sum of upward forces equals the weight.

3. Express this relationship using the variables P, C, and W.

Try solving on your own before revealing the answer!

Q13. A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the average braking force acting on the car?

Background

Topic: Newton's Second Law and Kinematics

This question tests your ability to relate force, mass, acceleration, and kinematic equations.

Key Formulas:

• Kinematic equation: $v^2 = v_0^2 + 2a d$

• Newton's Second Law: $F = m a$

Step-by-Step Guidance

1. Use the kinematic equation to solve for acceleration $a$ (final velocity is zero).

2. Plug in $v_0 = 27\ \mathrm{m/s}$, $v = 0$, and $d = 578\ \mathrm{m}$.

3. Once you have $a$, use $F = m a$ to find the braking force.

4. Remember to use the magnitude (ignore the negative sign for force direction).

Try solving on your own before revealing the answer!

Q14. A 55-kg box rests on a horizontal surface. The coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface?

Background

Topic: Frictional Forces

This question tests your understanding of static friction and the force required to initiate motion.

Key Formula:

$F_{\text{max}} = \mu_s N$

• $\mu_s$ = coefficient of static friction

• $N$ = normal force (equal to weight if surface is horizontal)

Step-by-Step Guidance

1. Calculate the normal force: $N = m g$.

2. Multiply by the coefficient of static friction to find the maximum static friction force.

3. The applied force must be just greater than this value to start motion.

Try solving on your own before revealing the answer!

Q15. The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. What is the force of friction on the box when a 15-N horizontal push is applied to the box?

Background

Topic: Frictional Forces

This question tests your ability to determine whether the box moves and which frictional force applies.

Key Formulas:

• Maximum static friction: $F_{s,\text{max}} = \mu_s N$

• Kinetic friction: $F_k = \mu_k N$

• $N = m g$

Step-by-Step Guidance

1. Calculate the normal force: $N = m g$.

2. Find the maximum static friction force.

3. Compare the applied force to the maximum static friction to determine if the box moves.

4. If the box moves, use kinetic friction; if not, use static friction.

Try solving on your own before revealing the answer!

Q16. An object slides on a level floor. It slows and comes to a stop with a constant acceleration of magnitude 2.4 m/s². What is the coefficient of kinetic friction between the object and the floor?

Background

Topic: Friction and Newton's Second Law

This question tests your ability to relate acceleration to the coefficient of kinetic friction.

Key Formula:

$f_k = \mu_k N$ and $F = m a$

• On a horizontal surface, $N = m g$

• Friction force causes the deceleration: $f_k = m a$

• So, $\mu_k = \frac{a}{g}$

Step-by-Step Guidance

1. Recognize that the only horizontal force is friction, which causes the acceleration.

2. Set $f_k = m a$ and $f_k = \mu_k m g$.

3. Set the two expressions for $f_k$ equal and solve for $\mu_k$.

4. Plug in the values for $a$ and $g$.

Try solving on your own before revealing the answer!

Q17. A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.35. What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s²?

Background

Topic: Newton's Second Law and Friction

This question tests your ability to calculate the net force required to overcome friction and produce a given acceleration.

Key Formulas:

• Kinetic friction: $f_k = \mu_k N$

• Newton's Second Law: $F_{\text{net}} = m a$

• Total applied force: $F_{\text{applied}} = f_k + F_{\text{net}}$

Step-by-Step Guidance

1. Calculate the normal force: $N = m g$.

2. Find the kinetic friction force: $f_k = \mu_k N$.

3. Calculate the net force needed for the desired acceleration: $F_{\text{net}} = m a$.

4. Add the friction force and the net force to find the total applied force required.

Try solving on your own before revealing the answer!