Physics with Calculus: Step-by-Step Guidance for Midterm Review

Q1. Which of the following is the dimensional formula for velocity?

Background

Topic: Dimensional Analysis

This question tests your understanding of how physical quantities are represented in terms of fundamental dimensions (length, mass, time, etc.). Velocity is a basic kinematic quantity.

Key Terms and Formulas:

• Velocity: The rate of change of displacement with respect to time.

• Dimensional formula: Expresses a physical quantity in terms of basic dimensions (L for length, M for mass, T for time).

• Velocity = Displacement / Time

Step-by-Step Guidance

1. Recall that displacement has the dimension of length $[L]$ and time has the dimension $[T]$.

2. Write the formula for velocity: $v = \frac{\text{displacement}}{\text{time}}$.

3. Express this in terms of dimensions: $[v] = \frac{[L]}{[T]}$.

4. Simplify the dimensional formula to match one of the options provided.

Try solving on your own before revealing the answer!

Final Answer: b. [L][T]^{-1}

The dimensional formula for velocity is $[L][T]^{-1}$, which matches option b.

Q2. An object's velocity-time graph shows a flat, horizontal line (not at zero) between $t=2$ s and $t=6$ s. Which of the following is true about the object during this time interval?

Background

Topic: Kinematics – Interpreting Graphs

This question tests your ability to interpret velocity-time graphs and understand what constant velocity implies about acceleration.

Key Terms and Concepts:

• Velocity-time graph: Shows how velocity changes with time.

• Flat, horizontal line: Indicates constant velocity.

• Acceleration: The rate of change of velocity with respect to time.

Step-by-Step Guidance

1. Recall that the slope of a velocity-time graph represents acceleration.

2. If the graph is a flat, horizontal line (not at zero), the velocity is constant and non-zero.

3. Consider what this means for acceleration: is it zero, constant, or changing?

4. Review the answer choices and match your reasoning to the correct statement.

Try solving on your own before revealing the answer!

Final Answer: c. The object has zero acceleration.

A flat, horizontal velocity-time graph means the velocity is constant, so acceleration (the slope) is zero.

Q3. A remote-control car has a velocity vector $\vec{v} = (6\hat{i} - 8\hat{j})$ m/s. What is the magnitude of the car's speed?

Background

Topic: Vectors – Magnitude of a Vector

This question tests your ability to calculate the magnitude of a vector given its components.

Key Terms and Formulas:

• Vector components: $v_x = 6$ m/s, $v_y = -8$ m/s

• Magnitude of a vector: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$

Step-by-Step Guidance

1. Identify the x and y components of the velocity vector: $v_x = 6$ m/s, $v_y = -8$ m/s.

2. Recall the formula for the magnitude of a vector in two dimensions: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$.

3. Substitute the given values into the formula.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: b. 10 m/s

$|\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ m/s.

Q4. A heavy box is being pushed across a horizontal floor at a constant speed in a straight line. Which statement about the forces acting on the box is true?

Background

Topic: Newton's Laws of Motion

This question tests your understanding of equilibrium and net force when an object moves at constant velocity.

Key Terms and Concepts:

• Constant speed: Implies zero acceleration.

• Net force: The vector sum of all forces acting on an object.

• Newton's First Law: An object in motion at constant velocity has zero net force.

Step-by-Step Guidance

1. Recall that constant speed means acceleration is zero.

2. Apply Newton's First Law: If acceleration is zero, the net force must also be zero.

3. Consider the forces acting: pushing force, friction, normal force, and weight.

4. Match this reasoning to the answer choices provided.

Try solving on your own before revealing the answer!

Final Answer: b. The net force on the box is zero.

Constant speed means no net force is acting on the box (forces are balanced).

Q5. A baseball is thrown straight upward. At the very peak of its flight, which of the following is true?

Background

Topic: Kinematics – Free Fall

This question tests your understanding of velocity and acceleration at the highest point of projectile motion.

Key Terms and Concepts:

• At the peak: The instant when the ball changes direction from upward to downward.

• Velocity: The rate of change of position; at the peak, vertical velocity is zero.

• Acceleration: Due to gravity, always $9.8$ m/s$^2$ downward near Earth's surface.

Step-by-Step Guidance

1. Recall that at the peak, the ball's vertical velocity is zero (it stops rising before falling back down).

2. However, gravity is still acting on the ball, so acceleration is not zero.

3. Identify the direction and magnitude of acceleration at the peak.

4. Compare your reasoning to the answer choices.

Try solving on your own before revealing the answer!

Final Answer: b. Its velocity is zero, and its acceleration is 9.8 m/s$^2$ downward.

At the peak, the ball's velocity is zero, but gravity still acts downward with $9.8$ m/s$^2$ acceleration.

Q6. Write $5 \times 10^4$ to 3 significant figures.

Background

Topic: Significant Figures and Scientific Notation

This question tests your ability to express numbers in scientific notation with the correct number of significant digits.

Key Terms and Concepts:

• Significant figures: The digits in a number that carry meaning contributing to its precision.

• Scientific notation: Expressing numbers as $a \times 10^n$ where $1 \leq a < 10$.

Step-by-Step Guidance

1. Express $5 \times 10^4$ in standard decimal form.

2. Determine how to write this number with three significant digits.

3. Rewrite the number in scientific notation, ensuring three significant figures are shown.

4. Check that the coefficient has exactly three digits.

Try solving on your own before revealing the answer!

Final Answer: $5.00 \times 10^4$

To show three significant figures, write $5.00 \times 10^4$.

Q7. A hiker walks in two stages. Their displacement vectors are $\vec{D}_1 = (2\hat{i} + 5\hat{j})$ m and $\vec{D}_2 = (3\hat{i} - 1\hat{j})$ m. (a) Determine the resultant displacement vector $\vec{D}_{\text{total}} = \vec{D}_1 + \vec{D}_2$. (b) Determine the vector difference $\vec{D}_1 - \vec{D}_2$. (c) Determine the scalar (dot) product $\vec{D}_1 \cdot \vec{D}_2$. (d) Determine the angle of the resultant displacement vector $\vec{D}_{\text{total}}$ relative to the positive x-axis.

Background

Topic: Vectors – Addition, Subtraction, Dot Product, and Direction

This problem tests your ability to perform vector operations and interpret the results geometrically.

Key Terms and Formulas:

• Vector addition: $\vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$

• Vector subtraction: $\vec{A} - \vec{B} = (A_x - B_x)\hat{i} + (A_y - B_y)\hat{j}$

• Dot product: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y$

• Angle with x-axis: $\theta = \arctan\left(\frac{A_y}{A_x}\right)$

Step-by-Step Guidance

1. For (a), add the corresponding components of $\vec{D}_1$ and $\vec{D}_2$ to find $\vec{D}_{\text{total}}$.

2. For (b), subtract the components of $\vec{D}_2$ from $\vec{D}_1$ to find the difference vector.

3. For (c), use the dot product formula with the given components.

4. For (d), use the components of $\vec{D}_{\text{total}}$ to set up the formula for the angle relative to the x-axis, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $\vec{D}_{\text{total}} = (5\hat{i} + 4\hat{j})$ m

• (b) $\vec{D}_1 - \vec{D}_2 = (-1\hat{i} + 6\hat{j})$ m

• (c) $\vec{D}_1 \cdot \vec{D}_2 = (2)(3) + (5)(-1) = 6 - 5 = 1$

• (d) $\theta = \arctan\left(\frac{4}{5}\right) \approx 38.7^\circ$

Each part uses the appropriate vector operation or formula as outlined above.

Q8. A stunt driver drives a car horizontally off a 20 m high cliff at a speed of 15 m/s. Air resistance is neglected. ($g = 9.8$ m/s$^2$) (a) How long is the car in the air before it hits the ground below? (b) How far horizontally from the base of the cliff does the car land? (c) What is the vertical component of the car's velocity just before it hits the ground? (d) What is the magnitude of the car's total velocity just before impact?

Background

Topic: Projectile Motion

This problem tests your understanding of two-dimensional kinematics, specifically horizontal projectile motion.

Key Terms and Formulas:

• Vertical motion: $y = y_0 + v_{y0} t - \frac{1}{2} g t^2$

• Horizontal motion: $x = v_{x0} t$

• Vertical velocity at impact: $v_{yf} = v_{y0} + g t$

• Total velocity: $v = \sqrt{v_x^2 + v_y^2}$

Step-by-Step Guidance

1. For (a), set up the vertical motion equation with $y_0 = 20$ m, $y = 0$, and $v_{y0} = 0$ (since the car drives off horizontally).

2. Solve for the time $t$ it takes to fall 20 m.

3. For (b), use the time from (a) and the horizontal velocity to set up the formula for horizontal distance.

4. For (c), use the time from (a) to find the vertical velocity just before impact.

5. For (d), set up the formula for the magnitude of the total velocity using the horizontal and vertical components.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $t = 2.02$ s

• (b) $x = 30.3$ m

• (c) $v_{yf} = 19.8$ m/s downward

• (d) $v = 24.7$ m/s

Each answer is found by applying the appropriate kinematic equation for projectile motion.

Q9. A car traveling horizontally at 25 m/s suddenly slams on the brakes. It comes to a complete stop in 3.5 s with constant deceleration. (a) What is the magnitude of the car's acceleration? (b) How far does the car travel while it is braking?

Background

Topic: Kinematics – Uniform Acceleration

This problem tests your ability to use kinematic equations to solve for acceleration and displacement during constant acceleration (deceleration).

Key Terms and Formulas:

• Initial velocity: $v_0 = 25$ m/s

• Final velocity: $v = 0$ m/s

• Time: $t = 3.5$ s

• Acceleration: $a = \frac{v - v_0}{t}$

• Displacement: $x = v_0 t + \frac{1}{2} a t^2$

Step-by-Step Guidance

1. For (a), use the acceleration formula with the given initial and final velocities and time.

2. Set up the calculation for acceleration, noting that it will be negative (deceleration).

3. For (b), use the displacement formula with the values for $v_0$, $t$, and $a$ (from part a).

4. Set up the calculation for displacement, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $a = -7.14$ m/s$^2$ (magnitude is $7.14$ m/s$^2$)

• (b) $x = 43.8$ m

Negative acceleration indicates the car is slowing down. The displacement is found using the kinematic equation.

Q10. A 40 kg box is on a frictionless horizontal floor. A person pulls on the box with a force of 150 N at an angle of 30$^\circ$ above the horizontal. ($g = 9.8$ m/s$^2$) (a) Draw the free body diagram of the box. (b) What is the horizontal acceleration of the box? (c) What is the Normal force acting on the box from the floor? (Hint: The rope is pulling slightly upwards, relieving some of the weight!)

Background

Topic: Newton's Second Law and Free Body Diagrams

This problem tests your ability to resolve forces into components, apply Newton's laws, and analyze normal force when an upward force is present.

Key Terms and Formulas:

• Force components: $F_x = F \cos \theta$, $F_y = F \sin \theta$

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

• Normal force: $N = mg - F_y$ (since the upward component of the pull reduces the normal force)

Step-by-Step Guidance

1. For (a), sketch the box and draw all forces: weight (down), normal force (up), pulling force (at 30$^\circ$ above horizontal), and resolve the pulling force into horizontal and vertical components.

2. For (b), calculate the horizontal component of the pulling force: $F_x = 150 \cos 30^\circ$.

3. Apply Newton's Second Law in the horizontal direction: $F_x = m a$ to solve for $a$.

4. For (c), calculate the vertical component of the pulling force: $F_y = 150 \sin 30^\circ$.

5. Set up the equation for the normal force: $N = mg - F_y$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) Free body diagram includes: weight ($mg$) down, normal force ($N$) up, pulling force at 30$^\circ$ above horizontal, resolved into $F_x$ and $F_y$.

• (b) $a = \frac{150 \cos 30^\circ}{40} = 3.25$ m/s$^2$

• (c) $N = mg - 150 \sin 30^\circ = 392 - 75 = 317$ N

The normal force is less than the weight because the rope pulls upward.

Q11. Suppose you launch a ball from a height of 4 m with an initial velocity of 5 m/s making a 0$^\circ$ angle with the horizontal. (a) How much time does it take to reach the floor? (b) How far horizontally does the ball reach on the floor?

Background

Topic: Projectile Motion (Horizontal Launch)

This problem tests your ability to analyze projectile motion when the initial vertical velocity is zero and the ball is launched horizontally from a height.

Key Terms and Formulas:

• Vertical motion: $y = y_0 + v_{y0} t - \frac{1}{2} g t^2$

• Horizontal motion: $x = v_{x0} t$

Step-by-Step Guidance

1. For (a), set up the vertical motion equation with $y_0 = 4$ m, $y = 0$, and $v_{y0} = 0$.

2. Solve for the time $t$ it takes to fall 4 m.

3. For (b), use the time from (a) and the horizontal velocity to set up the formula for horizontal distance.

4. Set up the calculation, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $t = 0.903$ s

• (b) $x = 4.51$ m

Time is found from the vertical motion equation, and horizontal distance is found using $x = v_{x0} t$.

Q12. A child of mass 20 kg rides on a toboggan down a slick, ice-covered hill inclined at an angle of 30$^\circ$ with respect to the horizontal. (a) Draw the Free body diagram of the child. (b) What is the acceleration of the child? (c) What is the normal force exerted on the child by the toboggan?

Background

Topic: Newton's Second Law on an Inclined Plane

This problem tests your ability to resolve forces on an incline and apply Newton's laws to find acceleration and normal force.

Key Terms and Formulas:

• Weight: $mg$

• Component of gravity parallel to incline: $mg \sin \theta$

• Component of gravity perpendicular to incline: $mg \cos \theta$

• Normal force: $N = mg \cos \theta$

• Acceleration down the incline: $a = g \sin \theta$

Step-by-Step Guidance

1. For (a), draw the child on the incline and show all forces: weight (down), normal force (perpendicular to surface), and resolve weight into components parallel and perpendicular to the incline.

2. For (b), use the formula $a = g \sin \theta$ to set up the calculation for acceleration.

3. For (c), use the formula $N = mg \cos \theta$ to set up the calculation for normal force.

4. Set up the calculations, but do not compute the final values yet.

Try solving on your own before revealing the answer!

Final Answers:

• (a) Free body diagram includes: weight ($mg$) down, normal force ($N$) perpendicular to incline, weight components $mg \sin \theta$ (down incline) and $mg \cos \theta$ (perpendicular to incline).

• (b) $a = 9.8 \sin 30^\circ = 4.9$ m/s$^2$

• (c) $N = 20 \times 9.8 \cos 30^\circ = 169.7$ N

Acceleration is due to the component of gravity down the incline; normal force is the perpendicular component.