Physics Exam Review Notes

Physics Exam Review Notes

1. Unit Conversion: Distance in Major League Baseball

• Problem Description: Convert the official pitching distance in Major League Baseball from feet and inches to meters.

• Given Values:- Pitching distance: $60 ext{ feet}, 6 ext{ inches}$

  • Conversion factors: $1 ext{ foot} = 12.000 ext{ inches}$

  • $1 ext{ inch} = 2.54 ext{ cm}$

• Step-by-Step Conversion:1. Convert feet to inches: Multiply the number of feet by the conversion factor for inches per foot.
  
  

  $60 ext{ ft} \times 12 ext{ in/ft} = 720 ext{ in}$

  2. Calculate total inches: Add the initial 6 inches to the converted feet.
     
     

     $720 ext{ in} + 6 ext{ in} = 726 ext{ in}$

  3. Convert total inches to centimeters: Multiply total inches by the conversion factor for centimeters per inch.
     
     

     $726 ext{ in} \times 2.54 ext{ cm/in} = 1844.04 ext{ cm}$

  4. Convert centimeters to meters: Divide the total centimeters by 100 (since $1 ext{ m} = 100 ext{ cm}$).
     
     

     $1844.04 ext{ cm} / 100 ext{ cm/m} = 18.4404 ext{ m}$

• Result: The pitching distance in meters is approximately $18.44 ext{ m}$.

2. Vector Angles

• Concept: Determining the angle of a 2D vector relative to the positive x-axis.

• Problem Description: Find the angle that a given vector makes with the positive x-axis.

• Given Vector: $\vec{A} = (30.0 ext{ m}, -50.0 ext{ m})$

• Methodology:1. Identify Components: The x-component is $A<em>x = 30.0 ext{ m}$ and the y-component is $A</em>y = -50.0 ext{ m}$.

  2. Use Arctangent Function: The angle $\theta$ can be found using the formula $\theta = \arctan(A<em>y / A</em>x)$.

  3. Calculate Raw Angle:
     
     

     $\theta_{raw} = \arctan(-50.0 / 30.0) \approx \arctan(-1.6667) \approx -59.0^[circ]$

  4. Adjust for Quadrant: Since $A<em>x$ is positive and $A</em>y$ is negative, the vector lies in the fourth quadrant. To express the angle as a positive value from the positive x-axis (counter-clockwise), add $360^[circ]$ to the raw calculated angle.
     
     

     $\theta = -59.0^[circ] + 360^[circ] = 301.0^[circ]$

• Result: The vector $\vec{A}$ makes an angle of approximately $301.0^[circ]$ with the positive x-axis.

3. Vector Components, Unit Vectors, Magnitude & Direction (General Method)

• Concept: Decomposing vectors into components, expressing them using unit vectors, and calculating the resultant's magnitude and direction.

• Problem Description: Find the components of given vectors, write them in terms of unit vectors, and then calculate the magnitude and direction of a resultant vector formed by specific vector operations. (Note: A figure showing vector directions is essential for numerical calculation but is not provided in the transcript.)

• Given Magnitudes:- $A = 75.0$

  • $B = 60.0$

  • $C = 25.0$

  • $D = 90.0$

• Resultant Vector Expression: $\vec{R} = \vec{A} - \vec{B} + \vec{D}$

• General Steps for Solution:1. Determine Components of Individual Vectors: For each vector (e.g., $\vec{V}$) with magnitude $\vert V\Vert$ and an angle $\theta$ (measured from the positive x-axis):- x-component: $V_x = \vert V\Vert \cos(\theta)$
  - y-component: $V_y = \vert V\Vert \sin(\theta)$

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  2. Write Vectors in Unit Vector Notation: Express each vector in terms of the unit vectors $[ ext{i}]\text{hat}$ (x-direction) and $[ ext{j}]\text{hat}$ (y-direction):- $\vec{V} = V<em>x \hat{i} + V</em>y \hat{j}$

  3. Calculate Components of the Resultant Vector ($\vec{R}$): Sum or subtract the x-components and y-components separately according to the given vector operation:- $R<em>x = A</em>x - B<em>x + D</em>x$

     • $R<em>y = A</em>y - B<em>y + D</em>y$

  4. Write Resultant Vector in Unit Vector Notation:- $\vec{R} = R<em>x \hat{i} + R</em>y \hat{j}$

  5. Calculate the Magnitude of the Resultant Vector: Use the Pythagorean theorem:- $\Vert\vec{R}\Vert = \sqrt{R<em>x^2 + R</em>y^2}$

  6. Calculate the Direction of the Resultant Vector: Use the arctangent function, adjusting for the correct quadrant based on the signs of $R<em>x$ and $R</em>y$:- $\phi = \arctan(R<em>y / R</em>x)$

4. Dot Product (Scalar Product)

• Concept: The dot product of two vectors, also known as the scalar product, yields a scalar quantity. It is defined as $\;\vec{A} \cdot \vec{B} = \vert A\Vert \vert B\Vert \cos(\phi)$, where $\phi$ is the angle between the vectors.

• Problem Description: Identify which of six cases (not provided) has the largest absolute value of the dot product $\;\vec{A} \cdot \vec{B}$, and then calculate it for that case.

• Analysis for Maximum Absolute Value:- The value of $\;\cos(\phi)$ ranges from $-1$ to $1$. Therefore, the absolute value $\vert\cos(\phi)\Vert$ ranges from $0$ to $1$.

  • $\Vert\vec{A} \cdot \vec{B}\Vert = \Vert\vert A\Vert \vert B\Vert \cos(\phi)\Vert = \vert A\Vert \vert B\Vert \Vert\cos(\phi)\Vert$

  • The maximum absolute value of the dot product occurs when $\Vert\cos(\phi)\Vert$ is maximized, which is $1$.

  • This happens when the angle $\phi$ is $0^[circ]$ (vectors are parallel) or $180^[circ]$ (vectors are anti-parallel). In these cases, the vectors are collinear.

• Calculation for the Selected Case (General):- If the vectors are parallel ($\phi = 0^[circ]$), then $\;\vec{A} \cdot \vec{B} = \vert A\Vert \vert B\Vert (1) = \vert A\Vert \vert B\Vert$.

  • If the vectors are anti-parallel ($\phi = 180^[circ]$), then $\;\vec{A} \cdot \vec{B} = \vert A\Vert \vert B\Vert (-1) = -\vert A\Vert \vert B\Vert$.

  • In both cases, the absolute value is $\vert A\Vert \vert B\Vert$. To find the specific numerical answer, one would multiply the magnitudes of vectors A and B from the figure (which is missing).

5. Cross Product (Vector Product)

• Concept: The cross product of two vectors, also known as the vector product, yields a vector quantity perpendicular to both original vectors. Its magnitude is defined as $\Vert\vec{A} \times \vec{B}\Vert = \vert A\Vert \vert B\Vert \sin(\phi)$, where $\phi$ is the angle between the vectors. The direction is determined by the right-hand rule.

• Problem Description: Identify which of six cases (not provided) has the largest absolute value of the cross product $\;\vec{A} \times \vec{B}$, and then calculate it for that case.

• Analysis for Maximum Absolute Value:- The value of $\;\sin(\phi)$ ranges from $-1$ to $1$. Therefore, the absolute value $\Vert\sin(\phi)\Vert$ ranges from $0$ to $1$.

  • The maximum absolute value of the cross product occurs when $\Vert\sin(\phi)\Vert$ is maximized, which is $1$.

  • This happens when the angle $\phi$ is $90^[circ]$ or $270^[circ]$ (vectors are perpendicular).

• Calculation for the Selected Case (General):- When the vectors are perpendicular ($\phi = 90^[circ]$ or $270^[circ]$), then $\Vert\sin(\phi)\Vert = 1$.

  • Thus, the magnitude of the cross product is $\Vert\vec{A} \times \vec{B}\Vert = \vert A\Vert \vert B\Vert (1) = \vert A\Vert \vert B\Vert$.

  • To find the specific numerical answer, one would multiply the magnitudes of vectors A and B from the figure (which is missing).

6. Rate of Continental Drift: Time Calculation

• Concept: Relationship between distance, rate, and time ($\text{distance} = \text{rate} \times \text{time}$) and extensive unit conversion.

• Problem Description: Estimate how long it took North America and Europe to reach their current separation, given the rate of continental drift.

• Given Values:- Rate of continental drift: $10 \text{ mm/yr}$

  • Current separation: $3000. \text{ miles}$

  • Conversion factor: $1 \text{ mile} = 1609 \text{ m}$

• Step-by-Step Calculation:1. Convert separation distance from miles to meters:
  
  

  $3000 \text{ miles} \times 1609 \text{ m/mile} = 4,827,000 \text{ m}$

  2. Convert separation distance from meters to millimeters: (To match the unit of the drift rate)
     
     

     $4,827,000 \text{ m} \times 1000 \text{ mm/m} = 4.827 \times 10^9 \text{ mm}$

  3. Calculate time: Use the formula $t = \text{distance} / \text{rate}$.
     
     

     $t = (4.827 \times 10^9 \text{ mm}) / (10 \text{ mm/yr}) = 4.827 \times 10^8 \text{ years}$

• Result: It took approximately $4.827 \times 10^8$ years (or 482.7 million years) for North America and Europe to reach their current separation.

7. Kinematics: Velocity-vs-Time Graph Analysis

• Concept: Interpreting velocity-vs-time (v-t) graphs to find average acceleration (slope) and total distance traveled (area under the curve).

• Problem Description: Given a v-t graph (not provided, but segmented into I, II, III), find the average acceleration during each segment and the total distance traveled over a specified time interval.

• General Methods:- a. Average Acceleration (Each Segment):- Average acceleration is the slope of the v-t graph for a given segment.
  - Formula: $a<em>{avg} = \Delta v / \Delta t = (v</em>f - v<em>i) / (t</em>f - t_i)$
  - For Segment I: Read the initial velocity ($v<em>0$) and final velocity ($v</em>1$) at times $t<em>0$ and $t</em>1$. Calculate $a<em>I = (v</em>1 - v<em>0) / (t</em>1 - t_0)$.
  - For Segment II: Read the initial velocity ($v<em>1$) and final velocity ($v</em>2$) at times $t<em>1$ and $t</em>2$. Calculate $a<em>{II} = (v</em>2 - v<em>1) / (t</em>2 - t_1)$. (If the segment is horizontal, velocity is constant, so acceleration is zero).
  - For Segment III: Read the initial velocity ($v<em>2$) and final velocity ($v</em>3$) at times $t<em>2$ and $t</em>3$. Calculate $a<em>{III} = (v</em>3 - v<em>2) / (t</em>3 - t_2)$.

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  • b. Total Distance Traveled ($t=0$ to $t=24 \text{ s}$):- The total distance traveled is the area under the v-t graph. This area can be calculated by dividing the region into simple geometric shapes (rectangles, triangles, or trapezoids) and summing their areas.

    • Area of a rectangle: $A = \text{base} \times \text{height}$

    • Area of a triangle: $A = 0.5 \times \text{base} \times \text{height}$

    • Area of a trapezoid: $A = 0.5 \times (\text{sum of parallel sides}) \times \text{height}$

    • Note: If the velocity is negative, the displacement is negative, but distance traveled is the absolute value of speed multiplied by time, or the sum of absolute areas. For typical kinematic problems where speed is non-negative, area under the curve is distance.

• Application: Without the specific graph, numerical calculation is not possible, but these are the established methods.

8. Kinematics: Variable Acceleration (Integration)

• Concept: Using integration to derive velocity from a time-dependent acceleration function and position from a time-dependent velocity function. This applies when acceleration is not constant.

• Problem Description: An object starts from rest with a given acceleration function, and we need to find its velocity and position after a certain time.

• Given Information:- Acceleration function: $a = Bt^2 - \frac{1}{2}Ct$

  • Given constants: $B = 2.0 \text{ m/s}^4$

  • $C = -4.0 \text{ m/s}^3$

  • Initial conditions: Starts from rest ($v<em>0 = 0$ at $t=0$). Assume initial position $x</em>0 = 0$ at $t=0$.

• Step-by-Step Solution:1. Substitute constants into the acceleration function:
  
  

  $a(t) = (2.0 \text{ m/s}^4)t^2 - \frac{1}{2}(-4.0 \text{ m/s}^3)t$
  
  

  $a(t) = 2.0t^2 + 2.0t \text{ (in m/s}^2 \text{ units)}$

  2. a. Find velocity as a function of time (integrate acceleration):
     
     

     $v(t) = \int a(t) dt = \int (2.0t^2 + 2.0t) dt = \frac{2.0}{3}t^3 + \frac{2.0}{2}t^2 + D<em>1$ Since the object starts from rest ($v(0) = 0$), the constant of integration $D</em>1 = 0$.
     
     

     $v(t) = \frac{2}{3}t^3 + t^2 \text{ (in m/s units)}$

  3. Calculate velocity after $5.0 \text{ s}$:
     
     

     $v(5.0 \text{ s}) = \frac{2}{3}(5.0)^3 + (5.0)^2 = \frac{2}{3}(125) + 25 = 83.333… + 25 = 108.33 \text{ m/s}$

  4. b. Find position as a function of time (integrate velocity):
     
     

     $x(t) = \int v(t) dt = \int (\frac{2}{3}t^3 + t^2) dt = \frac{2}{3 \times 4}t^4 + \frac{1}{3}t^3 + D<em>2$ $x(t) = \frac{1}{6}t^4 + \frac{1}{3}t^3 + D</em>2$
     
     

     Since we assume initial position is $0$ ($x(0) = 0$), the constant of integration $D_2 = 0$.

     $x(t) = \frac{1}{6}t^4 + \frac{1}{3}t^3 \text{ (in m units)}$

  5. Calculate distance moved after $5.0 \text{ s}$:
     
     

     $x(5.0 \text{ s}) = \frac{1}{6}(5.0)^4 + \frac{1}{3}(5.0)^3 = \frac{1}{6}(625) + \frac{1}{3}(125) = 104.166… + 41.666… = 145.83 \text{ m}$

• Results:- Velocity after $5.0 \text{ s}: \approx 108.3 \text{ m/s}$

  • Distance moved after $5.0 \text{ s}: \approx 145.8 \text{ m}$

9. Projectile Motion: Vertical Throw

• Concept: Kinematics under constant gravitational acceleration for vertically launched objects.

• Problem Description: A stone is thrown upward from ground level, and we need to determine its velocity and height after a specific time.

• Given Information:- Initial velocity: $v_0 = 10.0 \text{ m/s}$ (upward)

  • Initial position: $y_0 = 0$ (ground level)

  • Acceleration due to gravity: $a = -g = -9.8 \text{ m/s}^2$ (downward is negative)

• Step-by-Step Solution:1. a. Calculate velocity after $0.50 \text{ s}$: Use the kinematic equation $v = v_0 + at$
  
  

  $v = 10.0 \text{ m/s} + (-9.8 \text{ m/s}^2)(0.50 \text{ s})$
  
  

  $v = 10.0 - 4.9 = 5.1 \text{ m/s}$
  
  

  The positive value indicates the stone is still moving upward.

  2. b. Calculate height after $0.50 \text{ s}$: Use the kinematic equation $y = y<em>0 + v</em>0t + \frac{1}{2}at^2$
     
     

     $y = 0 + (10.0 \text{ m/s})(0.50 \text{ s}) + \frac{1}{2}(-9.8 \text{ m/s}^2)(0.50 \text{ s})^2$
     
     

     $y = 5.0 + \frac{1}{2}(-9.8)(0.25) = 5.0 - 1.225 = 3.775 \text{ m}$

• Results:- Velocity after $0.50 \text{ s}: 5.1 \text{ m/s}$ (upward)

  • Height after $0.50 \text{ s}: 3.775 \text{ m}$ above ground level

10. Kinematics: Stopping Distance and Time

• Concept: Applying kinematic equations for motion with constant deceleration to calculate stopping distance and time.

• Problem Description: A car traveling at a certain speed applies brakes with a constant deceleration to avoid hitting a truck ahead. We need to find how close it comes to the truck and the time taken to stop.

• Given Information:- Initial velocity: $v_0 = 29.1 \text{ m/s}$

  • Distance to truck: $d_{truck} = 200.0 \text{ m}$

  • Constant deceleration: $a = -2.4 \text{ m/s}^2$ (negative sign because it's slowing down)

  • Final velocity (when stopped): $v_f = 0$ (for stopping calculations)

• Step-by-Step Solution:1. b. Calculate time to come to a stop: Use the kinematic equation $v<em>f = v</em>0 + at$
  
  

  $0 = 29.1 \text{ m/s} + (-2.4 \text{ m/s}^2)t$
  
  

  $2.4t = 29.1$
  
  

  $t = 29.1 / 2.4 = 12.125 \text{ s}$

  2. a. Calculate stopping distance (x): Use the kinematic equation $v<em>f^2 = v</em>0^2 + 2ax$
     
     

     $0^2 = (29.1 \text{ m/s})^2 + 2(-2.4 \text{ m/s}^2)x$
     
     

     $0 = 846.81 - 4.8x$
     
     

     $4.8x = 846.81$
     
     

     $x = 846.81 / 4.8 = 176.41875 \text{ m}$

  3. Calculate how close the car comes to the truck's bumper: Subtract the stopping distance from the initial distance to the truck.
     
     

     $d<em>{close} = d</em>{truck} - x = 200.0 \text{ m} - 176.41875 \text{ m} = 23.58125 \text{ m}$

• Results:- The car comes approximately $23.58 \text{ m}$ close to the bumper of the truck.

  • It takes $12.125 \text{ s}$ for the car to come to a full stop.

11. Vector Addition: Sailboat Journey (Displacement)

• Concept: Vector addition to find a resultant displacement or an unknown vector component when the resultant and other vectors are known.

• Problem Description: A sailboat completes a journey in three legs. The first two legs and the final overall displacement are known. We need to find the magnitude and direction of the third, unknown leg.

• Given Displacements:- Leg 1 ($\vec{d_1}$): $2.00 \text{ km}$ East

  • Leg 2 ($\vec{d_2}$): $4.00 \text{ km}$ Southeast ($45^[circ]$ South of East)

  • Final Position (Resultant, $\vec{R}$): $6.00 \text{ km}$ directly East of the starting point

• Fundamental Principle: The sum of all individual displacement vectors equals the total (net) displacement: $\vec{R} = \vec{d<em>1} + \vec{d</em>2} + \vec{d_3}$

• Step-by-Step Solution:1. Resolve Known Vectors into x and y Components: (Assume East is positive x, North is positive y)

  • $\vec{d<em>1}$: $d</em>{1x} = 2.00 \text{ km}$, $d_{1y} = 0 \text{ km}$

  • $\vec{d_2}$: The angle for southeast is $-45^[circ]$ or $315^[circ]$ from the positive x-axis.- $d_{2x} = 4.00 \cos(-45^[circ]) = 4.00 (\sqrt{2}/2) \approx 2.828 \text{ km}$
    - $d_{2y} = 4.00 \sin(-45^[circ]) = 4.00 (-\sqrt{2}/2) \approx -2.828 \text{ km}$

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    • $\vec{R}$: $R<em>x = 6.00 \text{ km}$, $R</em>y = 0 \text{ km}$

  1. Set up Component Equations for the Resultant Vector:- $R<em>x = d</em>{1x} + d<em>{2x} + d</em>{3x}$

     • $R<em>y = d</em>{1y} + d<em>{2y} + d</em>{3y}$

  2. Solve for the Components of the Third Leg ($\vec{d_3}$):- For x-component: $6.00 = 2.00 + 2.828 + d<em>{3x}$ $d</em>{3x} = 6.00 - 2.00 - 2.828 = 1.172 \text{ km}$

     • For y-component: $0 = 0 - 2.828 + d<em>{3y}$ $d</em>{3y} = 2.828 \text{ km}$

  3. Calculate the Magnitude of the Third Leg: Use the Pythagorean theorem for its components.
     
     

     $\Vert\vec{d<em>3}\Vert = \sqrt{d</em>{3x}^2 + d<em>{3y}^2} = \sqrt{(1.172)^2 + (2.828)^2}$ $\Vert\vec{d</em>3}\Vert = \sqrt{1.3736 + 8.0000} = \sqrt{9.3736} \approx 3.062 \text{ km}$

  4. Calculate the Direction of the Third Leg: Use the arctangent function.
     
     

     $\theta = \arctan(d<em>{3y} / d</em>{3x}) = \arctan(2.828 / 1.172) \approx \arctan(2.413) \approx 67.5^[circ]$
     
     

     Since both $d<em>{3x}$ and $d</em>{3y}$ are positive, the vector is in the first quadrant, so the angle is North of East.

• Results: The third leg of the journey has a magnitude of approximately $3.062 \text{ km}$ and a direction of approximately $67.5^[circ]$ North of East.

12. Projectile Motion: Range Equation

• Concept: Using the horizontal range formula for projectile motion to find the initial speed required for a given range and launch angle (assuming launch and landing are on the same horizontal plane).

• Problem Description: Determine the initial speed needed for a catapult to throw rocks to a specific range at a given launch angle.

• Given Information:- Launch angle: $\theta = 45^[circ]$

  • Horizontal Range: $R = 0.67 \text{ km}$

  • Acceleration due to gravity: $g = 9.8 \text{ m/s}^2$

• Formula: The range equation is $R = (v_0^2 \sin(2\theta)) / g$

• Step-by-Step Solution:1. Convert Range to Meters:
  
  

  $R = 0.67 \text{ km} = 670 \text{ m}$

  2. Rearrange the Range Formula to Solve for Initial Speed (x):
     
     

     $v<em>0^2 = (R \cdot g) / \sin(2\theta)$ $v</em>0 = \sqrt{(R \cdot g) / \sin(2\theta)}$

  3. Substitute Given Values:- First calculate $2\theta = 2 \times 45^[circ] = 90^[circ]$

     • $\sin(90^[circ]) = 1$
       
       

       $v<em>0 = \sqrt{(670 \text{ m} \times 9.8 \text{ m/s}^2) / 1}$ $v</em>0 = \sqrt{6566} \approx 81.03 \text{ m/s}$

• Result: The rocks need an initial speed of approximately $81.03 \text{ m/s}$ to achieve the desired range.

13. Relative Velocity: Moving Walkway

• Concept: Calculating relative velocity in a one-dimensional scenario, specifically when velocities are collinear (in the same direction).

• Problem Description: A person walks on a moving walkway. Given the length of the walkway and the velocities of the person relative to the walkway and the walkway relative to the ground, calculate the time it takes for the person to traverse the walkway.

• Given Information:- Length of walkway: $L = 59.1 \text{ m}$

  • Velocity of person relative to walkway: $v_{pw} = 2.35 \text{ m/s}$

  • Velocity of walkway relative to ground: $v_{wg} = 1.77 \text{ m/s}$

• Fundamental Principle: The velocity of the person relative to the ground (the fixed reference frame) is the sum of their velocity relative to the walkway and the walkway's velocity relative to the ground: $\vec{v}<em>{pg} = \vec{v}</em>{pw} + \vec{v}_{wg}$.

• Step-by-Step Solution:1. Calculate the velocity of the person relative to the ground (effective speed): Since the person and walkway move in the same direction, their speeds add.
  
  

  $v_{pg} = 2.35 \text{ m/s} + 1.77 \text{ m/s} = 4.12 \text{ m/s}$

  2. Calculate the time taken: Use the formula $t = \text{distance} / \text{speed}$.
     
     

     $t = L / v_{pg} = 59.1 \text{ m} / 4.12 \text{ m/s} = 14.344… \text{ s}$

• Result: It will take approximately $14.34 \text{ s}$ to reach the other end of the walkway.

14. Relative Velocity: Raindrops from a Moving Car

• Concept: Calculating relative velocity in two dimensions, specifically to find the apparent velocity of an object as perceived from a moving reference frame. This involves vector subtraction.

• Problem Description: Rain falls vertically, and a car travels horizontally. We need to find the angle at which the raindrops appear to be falling to the driver of the car (i.e., the velocity of rain relative to the car).

• Given Information:- Velocity of rain relative to ground (vertical): $\vec{v}_{rg} = 7.00 \text{ m/s}$ (downward)

  • Speed of car relative to ground (horizontal): $v_{cg} = 60.0 \text{ km/h}$

• Fundamental Principle: The velocity of the rain relative to the car (observer) is given by $\vec{v}<em>{rc} = \vec{v}</em>{rg} - \vec{v}_{cg}$ (velocity of rain relative to ground minus velocity of car relative to ground).

• Step-by-Step Solution:1. Convert Car Speed to m/s:
  
  

  $v_{cg} = 60.0 \text{ km/h} \times (1000 \text{ m/km}) \times (1 \text{ h}/3600 \text{ s}) \approx 16.67 \text{ m/s}$

  2. Express Velocities in Component Form: (Let positive x be the direction of the car, and positive y be upward).- $\vec{v}_{rg} = (0 \text{ m/s})\hat{i} + (-7.00 \text{ m/s})\hat{j}$

     • $\vec{v}_{cg} = (16.67 \text{ m/s})\hat{i} + (0 \text{ m/s})\hat{j}$

  3. Calculate the Velocity of Rain Relative to the Car ($\vec{v}_{rc}$):- $v<em>{rc,x} = v</em>{rg,x} - v_{cg,x} = 0 - 16.67 = -16.67 \text{ m/s}$

     • $v<em>{rc,y} = v</em>{rg,y} - v<em>{cg,y} = -7.00 - 0 = -7.00 \text{ m/s}$ So, $\vec{v}</em>{rc} = (-16.67 \text{ m/s})\hat{i} + (-7.00 \text{ m/s})\hat{j}$.

  4. Find the Angle from the Vertical: The angle from the vertical is typically measured from the negative y-axis (towards the car's direction of travel or opposite). We'll find the angle $\alpha$ with the negative y-axis.
     
     

     $\tan(\alpha) = \vert v<em>{rc,x}\Vert / \vert v</em>{rc,y}\Vert = 16.67 / 7.00 \approx 2.3814$
     
     

     $\alpha = \arctan(2.3814) \approx 67.2^[circ]$
     
     

     The raindrops appear to be falling backward (relative to the car's motion) and downward.

• Result: The raindrops appear to be falling at an angle of approximately $67.2^[circ]$ from the vertical (towards the direction opposite to the car's motion).

15. Newton's Laws: Blocks in Contact

• Concept: Applying Newton's Second Law ($\;\vec{F}_ {net} = m\vec{a}$) to a system of multiple objects in contact. This involves using free-body diagrams and considering the system as a whole or individual components.

• Problem Description: Two blocks are in contact on a frictionless horizontal tabletop, and an external force is applied to block 1, causing both to accelerate together.

• Given Information:- Mass of block 1: $M_1 = 3.20 \text{ kg}$

  • Mass of block 2: $M_2 = 5.70 \text{ kg}$

  • Constant acceleration of both blocks: $a = 2.45 \text{ m/s}^2$

  • Frictionless surface (so no friction force acts on the blocks horizontally).

• Step-by-Step Solution:1. a. Calculate the Magnitude of the Applied Force F: Consider the two blocks as a single system. The total mass is $M<em>{total} = M</em>1 + M_2$. The external force F is responsible for accelerating this total mass.- $M_{total} = 3.20 \text{ kg} + 5.70 \text{ kg} = 8.90 \text{ kg}$
  - Applying Newton's Second Law to the system: $F = M_{total} \cdot a$
  - $F = 8.90 \text{ kg} \times 2.45 \text{ m/s}^2 = 21.805 \text{ N}$

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  2. b. Calculate the Net Force Acting on Block 1: The net force on any object is given by its mass times its acceleration. This is a direct application of Newton's Second Law specifically for Block 1.- $\;\vec{F}<em>{net,1} = M</em>1 \cdot \vec{a}$

     • Since only magnitude is requested: $F<em>{net,1} = M</em>1 \cdot a = 3.20 \text{ kg} \times 2.45 \text{ m/s}^2 = 7.84 \text{ N}$
       
       

     (Note: The actual forces acting on Block 1 are the applied force F and the normal contact force from Block 2, $F</em>{21}$. The net force is $F - F<em>{21}$, which equals $M_1a$. This matches the calculated value.)

• Results:- The magnitude of the applied force F is approximately $21.81 \text{ N}$.

  • The net force acting on block 1 is $7.84 \text{ N}$.

16. Friction and Kinematics: Stopping a Hockey Puck

• Concept: Combining kinematic equations (to find acceleration) with Newton's Second Law and the definition of kinetic friction to determine the coefficient of friction.

• Problem Description: A hockey puck slides to a stop over a certain distance after being given an initial speed. We need to find the coefficient of friction required for this to happen.

• Given Information:- Initial speed: $v_0 = 12.5 \text{ m/s}$

  • Stopping distance: $x = 60.5 \text{ m}$

  • Final speed: $v_f = 0$ (since it stops)

• Step-by-Step Solution:1. Calculate the Acceleration (Deceleration): Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
  
  

  $v<em>f^2 = v</em>0^2 + 2ax$
  
  

  $0^2 = (12.5 \text{ m/s})^2 + 2a(60.5 \text{ m})$
  
  

  $0 = 156.25 + 121a$
  
  

  $121a = -156.25$
  
  

  $a = -156.25 / 121 \approx -1.29132 \text{ m/s}^2$ (The negative sign indicates deceleration, as expected).

  2. Apply Newton's Second Law:- Vertical Forces: The normal force ($N$) balances the gravitational force ($mg$) since there is no vertical acceleration.
     
     

     $N - mg = 0 \implies N = mg$

     • Horizontal Forces: The only horizontal force acting on the puck is the kinetic friction force ($f<em>k$), which opposes the motion and causes the deceleration. According to Newton's Second Law, $F</em>{net} = ma$.
       
       

       $f_k = m\vert a\Vert$ (We use the magnitude of acceleration, as friction itself is a magnitude, and we typically assign signs for direction in a coordinate system).

  3. Relate Friction Force to Coefficient of Friction: The kinetic friction force is given by $f<em>k = \mu</em>k N$.

  4. Equate and Solve for $\;\mu<em>k$:*
     
     

     $\mu*k N = m\vert a\Vert$
     
     

     Substitute $N = mg$
     
     

     $\mu<em>k mg = m\vert a\Vert$ Cancel mass (m) from both sides (demonstrates that the coefficient of friction is independent of mass): $\mu</em>k g = \vert a\Vert$
     
     

     $\mu<em>k = \vert a\Vert / g = 1.29132 \text{ m/s}^2 / 9.8 \text{ m/s}^2$ $\mu</em>k \approx 0.13176$

• Result: A coefficient of friction of approximately $0.132$ is required to stop the puck under these conditions.

17. Rotational Motion and Unit Conversion: Baseball Spin

• Concept: Calculating the number of rotations an object makes during a certain linear travel time, requiring conversion between linear speed, rotational speed, distance, and time.

• Problem Description: A baseball is pitched at a given speed and spin rate, covering a specified distance. We need to find how many full turns the ball makes during its flight.

• Given Information:- Baseball speed: $v = 88.0 \text{ mph}$

  • Spin rate: $\omega = 110. \text{ rpm}$

  • Distance between release and catch: $d = 60.5 \text{ ft}$

  • Neglect gravity or air resistance for determining flight time.

• Step-by-Step Solution:1. Convert Linear Speed to feet per second (ft/s):
  
  

  $v = 88.0 \frac{\text{miles}}{\text{hour}} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ s}} \approx 129.0667 \text{ ft/s}$

  2. Convert Spin Rate to revolutions per second (rps):
     
     

     $\omega = 110 \frac{\text{rev}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} \approx 1.8333 \text{ rps}$

  3. Calculate the Time of Flight: Use the formula $t = d / v$.
     
     

     $t = 60.5 \text{ ft} / 129.0667 \text{ ft/s} \approx 0.46875 \text{ s}$

  4. Calculate the Total Number of Turns: Multiply the spin rate (in rps) by the time of flight.
     
     

     $\text{Total Turns} = \omega \times t = 1.8333 \text{ rps} \times 0.46875 \text{ s} \approx 0.859375 \text{ revolutions}$

  5. Determine the Number of Full Turns: Since the total turns is less than 1, the number of full turns is 0.

• Result: The ball makes 0 full turns between release and catch based on the given values.

18. Uniform Circular Motion: Centripetal Force

• Concept: Calculating centripetal force, which is the net force required to keep an object moving in a circular path. Also, identifying the physical force that provides this centripetal force for a given scenario.

• Problem Description: Calculate the centripetal force on a car moving around a curve and identify the physical force responsible for it.

• Given Information:- Mass of vehicle: $m = 1500 \text{ kg}$

  • Speed of vehicle: $v = 15.0 \text{ m/s}$

  • Radius of curve: $R = 400. \text{ m}$

• Fundamental Principle: Centripetal force (directed towards the center of the circular path) is given by $F_c = mv^2 / R$.

• Step-by-Step Solution:1. Calculate the Centripetal Force:
  
  

  $F<em>c = (1500 \text{ kg} \times (15.0 \text{ m/s})^2) / 400. \text{ m}$ $F</em>c = (1500 \text{ kg} \times 225 \text{ m}^2\text{/s}^2) / 400. \text{ m}$
  
  

  $F_c = 337500 / 400 = 843.75 \text{ N}$

  2. Identify the Force Playing the Role of Centripetal Force: For a car moving on a flat, unbanked circular curve, the centripetal force is provided by the static friction between the car's tires and the road surface. This static friction prevents the car from sliding outwards (skidding) as it turns.

• Result:- The centripetal force exerted on the vehicle is $843.75 \text{ N}$.

  • The static friction force between the tires and the road plays the role of the centripetal force.

19. General Problem-Solving Strategy in Physics

To effectively approach physics problems, follow these systematic steps, which are implicitly demonstrated in the solutions provided:

1. Understand the Concept: Before diving into numbers, ensure you grasp the underlying physics principles and concepts relevant to the problem. What physical laws or definitions apply here?

2. Read and Visualize: Carefully read the problem statement. Draw a diagram, sketch a free-body diagram, or visualize the scenario. Label known quantities and variables that need to be found.

3. Identify Given Information and Unknowns: List all numerical values provided, along with their units. Clearly state what quantity (or quantities) the problem asks you to determine.

4. Perform Unit Conversions (if necessary): Convert all given values to a consistent set of units (e.g., SI units like meters, kilograms, seconds) before plugging them into equations.

5. Choose Appropriate Formulas/Principles: Based on the concepts and given information, select the relevant physics equations, laws, or theorems.

6. Set Up Equations: Write down the chosen formulas and substitute known values (after conversion). If there are multiple unknowns, you might need a system of equations.

7. Solve Algebraically First: Before substituting numbers, try to isolate the unknown variable algebraically. This helps avoid calculation errors and makes the solution reusable for different values.

8. Calculate the Numerical Answer: Plug in the numerical values into the rearranged formula and compute the result. Pay attention to significant figures and rounding.

9. Check Units: Ensure the units of your final answer are consistent with the quantity you calculated. This is a common way to catch errors.

10. Evaluate Reasonableness: Does your answer make sense in the context of the problem? Is the magnitude realistic? Does the direction (for vectors) match your expectation?

20. Physics Formulas Cheat Sheet

20.1. Unit Conversions

• Distance: $1 \text{ foot} = 12.000 \text{ inches}$

• Distance: $1 \text{ inch} = 2.54 \text{ cm}$

• Distance: $1 \text{ m} = 100 \text{ cm}$

• Distance: $1 \text{ mile} = 1609 \text{ m}$

20.2. Kinematics (Constant Acceleration)

• Average Acceleration: $a<em>{avg} = \Delta v / \Delta t = (v</em>f - v<em>i) / (t</em>f - t_i)$

• Velocity: $v = v_0 + at$

• Position: $y = y<em>0 + v</em>0t + \frac{1}{2}at^2$

• Velocity-Displacement: $v<em>f^2 = v</em>0^2 + 2ax$

20.3. Kinematics (Variable Acceleration)

• Velocity from Acceleration: $v(t) = \int a(t) dt$

• Position from Velocity: $x(t) = \int v(t) dt$

20.4. Projectile Motion

• Horizontal Range (same level): $R = (v_0^2 \sin(2\theta)) / g$

20.5. Vectors

• Vector Angle (from components): $\theta = \arctan(A<em>y / A</em>x)$ (adjust for quadrant)

• x-component: $V_x = \vert V\Vert \cos(\theta)$

• y-component: $V_y = \vert V\Vert \sin(\theta)$

• Vector in Unit Notation: $\vec{V} = V<em>x \hat{i} + V</em>y \hat{j}$

• Resultant Magnitude: $\Vert\vec{R}\Vert = \sqrt{R<em>x^2 + R</em>y^2}$

• Resultant Direction: $\phi = \arctan(R<em>y / R</em>x)$ (adjust for quadrant)

• Dot Product: $\;\vec{A} \cdot \vec{B} = \vert A\Vert \vert B\Vert \cos(\phi)$

• Cross Product Magnitude: $\Vert\vec{A} \times \vec{B}\Vert = \vert A\Vert \vert B\Vert \sin(\phi)$

20.6. Newton's Laws & Forces

• Newton's Second Law: $\;\vec{F}_ {net} = m\vec{a}$

• Kinetic Friction: $f<em>k = \mu</em>k N$

20.7. Relative Velocity

• General (1D): $\vec{v}<em>{pg} = \vec{v}</em>{pw} + \vec{v}_{wg}$

• General (2D): $\vec{v}<em>{rc} = \vec{v}</em>{rg} - \vec{v}_{cg}$

20.8. Uniform Circular Motion

• Centripetal Force: $F_c = mv^2 / R$

20.