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Question 1

A student has a cart of mass \( m_c \) and a block of mass \(\frac{1}{5}m_c\), as shown in Figure 1.
  • At time \( t = 0 \), the cart is moving to the right across a horizontal surface with constant speed \( v_c \), and the student releases the block from rest.
  • At \( t = t_1 \), the block collides with and sticks to the top of the cart. The block does not slide on the cart.
  • At \( t = t_2 \), the block-cart system continues to move to the right with constant speed \( v_f \).
 
 
 
 
 
 
 
 
 
 
A.
i. On the axes provided in Figure 2, sketch a graph of the magnitude \( p_x \) of the x-component of the momentum of the block-cart system as a function of time \( t \) from \( t = 0 \) until \( t > t_2 \).
 
 
 
 
 
 
 
 
 
 
 
 
ii. Derive an expression for the speed \( v_f \) of the block-cart system after time \( t = t_2 \) in terms of \( m_c, v_c \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
iii. Derive an expression for the change in the kinetic energy \( \Delta K \) in the block-cart system from \( t = 0 \) to \( t = t_2 \) in terms of \( m_c, v_c \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
B.
Consider the case where a new block is dropped and collides with the top of the cart. The new block slides along the cart during the collision but does not slide off the cart. The time interval from when the new block collides with the cart and moves together with the cart is \( \Delta t \). During \( \Delta t \) there is a frictional force between the new block and the cart.
Indicate whether the \( x \)-component of the momentum of the new block-cart system increases, decreases, or remains constant during \( \Delta t \).
□ Increases
□ Decreases
□ Remains constant
Justify your response.

Most-appropriate topic codes (AP Physics 1):

• 3.D.1.1: Linear momentum and impulse — parts A(i), A(ii), B
• 3.D.2.1: Conservation of linear momentum — parts A(ii), B
• 5.B.4.1: Kinetic energy — part A(iii)
• 5.B.5.1: Inelastic collisions and energy changes — part A(iii)
• 3.A.2.1: Forces as interactions — part B justification
▶️ Answer/Explanation

(A)(i)
For \( t < t_1 \): Only the cart is moving horizontally with speed \( v_c \), so \( p_x = m_c v_c \).
For \( t > t_1 \): Block and cart stick together; momentum is conserved: \( m_c v_c = \left(m_c + \frac{1}{5}m_c\right) v_f \), so \( p_x = m_c v_c \) remains constant.
✅ Answer: A constant, nonzero, horizontal line from \( t = 0 \) to \( t > t_2 \).

 

 

 

 

 

 

 

 

(A)(ii)
Start with conservation of momentum:
\( p_i = p_f \)
\( m_c v_c = \left(m_c + \frac{1}{5}m_c\right) v_f \)
\( m_c v_c = \frac{6}{5}m_c v_f \)
\( v_f = \frac{5}{6} v_c \)
✅ Answer: \( \boxed{v_f = \frac{5}{6} v_c} \).

(A)(iii)
Change in kinetic energy: \( \Delta K = K_f – K_i \)
\( K_i = \frac{1}{2} m_c v_c^2 \)
\( K_f = \frac{1}{2} \left(\frac{6}{5} m_c\right) \left(\frac{5}{6} v_c\right)^2 = \frac{5}{12} m_c v_c^2 \)
\( \Delta K = \frac{5}{12} m_c v_c^2 – \frac{1}{2} m_c v_c^2 = -\frac{1}{12} m_c v_c^2 \)
✅ Answer: \( \boxed{\Delta K = -\frac{1}{12} m_c v_c^2} \).

(B)
The frictional force between block and cart is internal to the block-cart system. No net external force acts on the system in the horizontal direction, so the horizontal momentum remains constant.
✅ Answer: Remains constant.
Justification: The friction is an internal force; only external forces can change the total momentum of a system. Here, the vertical forces are balanced, and no horizontal external force acts.

Question 2

A block of mass \( M \) is released from rest at position \( x = 0 \) near the top of a ramp. The ramp makes an angle \( \theta \) with the horizontal. The block slides with negligible friction. At \( x = 8D \) it contacts an uncompressed spring (spring constant \( k \)). The spring compresses until the block momentarily stops at \( x = 12D \).

(a) Figure 4 (see exam) gives an energy bar chart for \( x = 10D \), with gravitational potential energy \( U_g \) defined as zero at \( x = 12D \). Complete the energy bar charts in Figure 2 (for \( x = 0 \)) and Figure 3 (for \( x = 6D \)) by drawing shaded bars for kinetic energy \( K \), gravitational potential energy \( U_g \), and spring potential energy \( U_s \).
Draw shaded bars that represent K,\(U_g\), and\(U_s\)to complete the energy bar charts in Figure 2 and Figure 3 for when the block is released from rest at x=0 and for when the block is at x = D6, respectively.
• Shaded bars should start at the dashed line that represents zero energy.
• Represent any energy that is equal to zero with a distinct line on the zero-energy line.
• The relative heights of each shaded bar should reflect the magnitude of the respective
energy consistent with the scale used in Figure 4.
 
 
 
 
 
 
 
 
 
 
 
(b) Figure 5 shows the block at x= 0 when the block is released from rest and the block at x = D12
when the block momentarily comes to rest against the compressed spring.
 
 
 
 
 
 
 
 
Using conservation of energy, derive an expression for the spring constant \( k \) in terms of \( M, \theta, D, g \).
(c) Figure 6 (see exam) shows \( U_s \) from \( x = 8D \) to \( 12D \). On the same axes:  (i) Sketch and label the total mechanical energy \( E \).  (ii) Sketch and label \( U_g \).
 
 
 
 
 
 
 
 
 
 
 
 
 
(d) Compare the speed of the block at \( x = 9D \) (\( v_{9D} \)) with its speed at \( x = 8D \) (\( v_{8D} \)). Justify using your energy graphs from part (c).

Most-appropriate topic codes (AP Physics 1 Course Framework):

• 3.E.1.1: Make predictions about the changes in kinetic and potential energy of a system – part (a)
• 4.C.1.1: Calculate the total energy of a system – parts (a), (c)
• 4.C.1.2: Predict changes in the total energy of a system – parts (a), (d)
• 5.B.3.1: Derive an expression for the gravitational potential energy of a system – part (b)
• 5.B.3.2: Derive an expression for the elastic potential energy of a system – part (b)
• 5.B.4.1: Describe and make predictions about the internal energy of systems – part (c)
• 5.B.5.4: Calculate energy changes in processes involving springs – part (b)
• 5.B.5.5: Make calculations of quantities related to energy of a system – part (b)
• 6.B.1.1: Design an experiment for collecting data to determine gravitational potential energy – implied in experimental context
▶️ Answer/Explanation

(a) Energy Bar Charts

  • At \( x = 0 \): Block is at rest (\( K = 0 \)), spring uncompressed (\( U_s = 0 \)), so all energy is gravitational: \( U_g = Mg(12D \sin\theta) = 12E_0 \) (scaled). Only one bar for \( U_g \).
  • At \( x = 6D \): Block has descended \( 6D\sin\theta \), so \( U_g = Mg(6D\sin\theta) = 6E_0 \). It has gained kinetic energy: \( K = 6E_0 \). \( U_s = 0 \). Two bars: \( K \) and \( U_g \), each of height \( 6E_0 \).
  • Total height of bars in each chart = \( 12E_0 \) (conservation of energy).

Correct charts:

 

 

 

 

 

 

 

 

(b) Derivation of \( k \)

Step 1: Write conservation of energy between \( x = 0 \) and \( x = 12D \):
\( U_{g,i} + K_i + U_{s,i} = U_{g,f} + K_f + U_{s,f} \)

Step 2: Substitute known values:

  • At \( x = 0 \): \( U_{g,i} = Mg(12D\sin\theta) \), \( K_i = 0 \), \( U_{s,i} = 0 \)
  • At \( x = 12D \): \( U_{g,f} = 0 \) (defined), \( K_f = 0 \) (momentarily at rest), \( U_{s,f} = \frac{1}{2}k(\Delta x)^2 \) with \( \Delta x = 4D \)

Step 3: Equation becomes:
\( Mg(12D\sin\theta) = \frac{1}{2}k(4D)^2 \)

Step 4: Solve for \( k \):
\( 12MgD\sin\theta = \frac{1}{2}k(16D^2) \)
\( 12MgD\sin\theta = 8kD^2 \)
\( k = \frac{12Mg\sin\theta}{8D} = \frac{3Mg\sin\theta}{2D} \)

✅ Final expression: \( \boxed{k = \frac{3Mg\sin\theta}{2D}} \)

(c) Energy Graphs from \( x = 8D \) to \( 12D \)

  • Total mechanical energy \( E \): Constant (no non‑conservative work), horizontal line at \( 12E_0 \).
  • Gravitational potential energy \( U_g \): Decreases linearly with \( x \) because \( U_g = Mg(y) \) and \( y = (12D – x)\sin\theta \). At \( x = 8D \), \( U_g = Mg(4D\sin\theta) = 4E_0 \); at \( x = 12D \), \( U_g = 0 \). Straight line from \( (8D, 4E_0) \) to \( (12D, 0) \).

 

 

 

 

 

 

 

(d) Speed Comparison

Claim: \( v_{9D} > v_{8D} \)

Justification:
From energy conservation: \( K = E – (U_g + U_s) \).
At \( x = 8D \): \( U_g = 4E_0 \), \( U_s = 0 \) ⇒ \( K_{8D} = 12E_0 – 4E_0 = 8E_0 \).
At \( x = 9D \): From the graph, \( U_g \approx 3E_0 \) and \( U_s < E_0 \) (since \( U_s \) curve is below \( E_0 \) at 9D).
Thus \( U_g + U_s < 4E_0 \) ⇒ \( K_{9D} > 8E_0 \).
Since \( K = \frac{1}{2}Mv^2 \) and \( M \) is constant, \( K_{9D} > K_{8D} \) ⇒ \( v_{9D} > v_{8D} \).

✅ Answer: \( \boxed{v_{9D} > v_{8D}} \) with justification referencing energy graph.

 

Question 3

 Students are investigating balancing systems using the setup shown in Figure 1. They have a spring scale of negligible mass attached to one end of a uniform meterstick. The meterstick’s center is attached to a stand on which the meterstick can pivot. A hook of negligible mass is fixed to the top of a block of unknown mass \( m_0 \). The hook can be attached to the meterstick through one of the small holes in the meterstick. The students do not have a direct way to measure the mass of the block, and the block cannot be attached to the spring scale.
 
 
 
 
 
 
 
 
 
 
 
 
The students are asked to take measurements that will allow the students to create a linear graph whose slope could be used to determine the mass \(m_0\) of the block.
A. Describe an experimental procedure to collect data that would allow the students to determine \( m_0 \). Include any steps necessary to reduce experimental uncertainty. 
B.Describe how the data collected in part A could be graphed and how that graph would be analyzed to determine \( m_0 \).
 Later, the students have an identical meterstick of mass \( M \) attached to an axle fixed to a wall. The meterstick is suspended horizontally by a string connected to a spring scale (Figure 2).
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
The angle \( \theta \) between the string and the meterstick can be varied. The relationship between the tension \( F_T \) and \( \theta \) is given as: \[ F_T = \frac{5Mg}{6\sin\theta}. \]
 
 
 
 
 
 
 
 
 
 
C. (i) Indicate what measured or calculated quantity could be plotted on the vertical axis (with \( \frac{1}{\sin\theta} \) on the horizontal axis) to yield a linear graph whose slope can be used to calculate an experimental value for \( M \).
(ii) On the provided grid, create a graph of the quantities indicated in part C(i). Label the vertical axis, include units, and plot the data points from Table 1.
 
 
 
 
 
 
 
 
 
 
 
 
 
(iii) Draw a straight best‑fit line for the data graphed in part C(ii).
D. Using the best‑fit line drawn in part C(iii), calculate an experimental value for the mass \( M \) of the meterstick.

Most-appropriate topic codes (AP Physics 1):

• 3.F.1.1: Torque and rotational statics — Part A, Part B
• 3.F.1.2: Conditions for equilibrium — Part A, Part B
• 5.B.9.1: Graphical analysis and linearization — Part C
• 5.B.10.1: Experimental design and uncertainty — Part A
• 5.B.10.2: Data collection and analysis — Part B, Part C, Part D
▶️ Answer/Explanation

A: Describes a procedure that includes:

    • Attaching the block to the meterstick.
    • Measuring the force exerted on the left end of the meterstick (spring‑scale reading) while the block is attached and the system is balanced horizontally.
    • Indicates a reasonable method to reduce experimental uncertainty (e.g., repeating measurements at one block location, or taking measurements with the block at multiple different hole positions).

B:

  • B1: Indicates appropriate quantities that can be plotted so that \( m_0 \) can be determined from the slope (e.g., spring‑scale force \( F \) vs. distance \( d \) from pivot to block, or \( F \) vs. \( d \cdot g \), etc.).
  • B2 : Describes how the slope of the graph is related to \( m_0 \).

C:

  • C1 : Lists a quantity for the vertical axis that yields a linear graph with slope related to \( M \) (e.g., \( F_T \), \( F_T/g \), \( \frac{6F_T}{5g} \), etc.).
  • C2 : Correctly labels the vertical axis (with units if appropriate) and uses a linear scale.
  • C3 : Plots data points consistent with the quantities indicated in C(i) or in Table 2.
  • C4 : Draws a reasonable straight best‑fit line that approximates the trend of the plotted data.

 

 

 

 

 

 

 

 

 

D:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D1 : Correctly relates the slope of the best‑fit line to \( M \) using the given equation \( F_T = \frac{5Mg}{6\sin\theta} \).

  • D2 : Calculates a value of \( M \) within the acceptable range (0.90 kg to 1.15 kg).    

Key Physics Principles and Methods

1. Torque Equilibrium (Parts A & B)
For the meterstick balanced horizontally about the pivot (stand), the net torque is zero: \[ \tau_{\text{spring scale}} = \tau_{\text{block}}. \] If the spring scale is attached at a fixed distance \( L_s \) from the pivot and the block is at distance \( d \) from the pivot, then \[ F \cdot L_s = m_0 g \cdot d, \] where \( F \) is the spring‑scale reading. Rearranging gives \[ F = \left( \frac{m_0 g}{L_s} \right) d. \] Thus, a graph of \( F \) vs. \( d \) is linear with slope \( \frac{m_0 g}{L_s} \), allowing \( m_0 \) to be determined if \( L_s \) and \( g \) are known.

2. Linearization and Graphical Analysis (Part C)
Given \( F_T = \frac{5Mg}{6\sin\theta} \), rewriting as \[ F_T = \left( \frac{5Mg}{6} \right) \frac{1}{\sin\theta} \] shows that plotting \( F_T \) (vertical) vs. \( \frac{1}{\sin\theta} \) (horizontal) yields a straight line through the origin with slope \( \frac{5Mg}{6} \). Alternative linear forms (e.g., \( \frac{F_T}{g} \) vs. \( \frac{1}{\sin\theta} \)) are also acceptable.

3. Uncertainty Reduction (Part A)
Acceptable methods include: • Taking multiple trials at each block position and averaging. • Using multiple different block positions to obtain several data points for the graph, which reduces the impact of measurement errors.

4. Calculation of \( M \) (Part D)
Using the best‑fit line slope \( k \) from the graph: • If \( F_T \) vs. \( \frac{1}{\sin\theta} \) is plotted, \( k = \frac{5Mg}{6} \), so \( M = \frac{6k}{5g} \). • Using typical values from the data (e.g., slope ≈ 8 N) and \( g = 9.8 \) m/s² gives \( M ≈ 0.98 \) kg, which lies within the acceptable range (0.90–1.15 kg).

Question 4 

In Scenario 1, a swimmer holds a block of mass \( m \) at rest in freshwater (density \( \rho_1 \)). The block is released and accelerates upward with initial acceleration \( a_1 \).
In Scenario 2, the same block is held at rest in salt water (density \( \rho_2 \), where \( \rho_2 > \rho_1 \)). The block is released and accelerates upward with initial acceleration \( a_2 \).
Frictional forces are negligible in both scenarios.
A. Indicate whether \( a_1 > a_2 \), \( a_1 < a_2 \), or \( a_1 = a_2 \). Justify in terms of all forces on the block.
B. For a block of mass \( m \) and volume \( V \) submerged in fluid of density \( \rho \), derive an expression for the initial upward acceleration \( a \) using Newton’s second law.
C. Indicate whether your derived expression in part B is consistent with your claim in part A, and justify.

Most-appropriate topic codes (AP Physics 1 Course and Exam Description):

• 3.A.1.1: Express the motion of an object using narrative or mathematical representations — part A justification
• 3.B.1.1: Predict the motion of an object subject to forces exerted by several objects using Newton’s second law — parts A, B
• 3.B.1.3: Reexpress a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object — part B derivation
• 4.C.1.1: Calculate the gravitational and buoyant forces on a submerged or floating object — parts A, B
• 5.D.1.6: Make claims about the behavior of physical systems based on conservation laws — part C consistency check
▶️ Answer/Explanation

(a)
The correct relationship is \( a_1 < a_2 \).
Justification:
• The gravitational force \( F_g = mg \) is identical in both scenarios because the block’s mass is the same.
• The buoyant force is given by \( F_B = \rho V g \), where \( V \) is the volume of the displaced fluid (equal to the block’s volume).
• Since \( \rho_2 > \rho_1 \), the buoyant force is larger in salt water (\( F_{B2} > F_{B1} \)).
• The net upward force is \( F_{\text{net}} = F_B – mg \).
• With a larger buoyant force in Scenario 2, the net upward force is greater, leading to a larger upward acceleration.
✅ Answer: \( \boxed{a_1 < a_2} \)

(b)
Start with Newton’s second law:
\( \sum F_y = m a \)
The forces are buoyant force upward and gravitational force downward:
\( F_B – mg = m a \)
Substitute \( F_B = \rho V g \):
\( \rho V g – mg = m a \)
Solve for \( a \):
\( a = \frac{\rho V g – mg}{m} = \frac{\rho V g}{m} – g \)
✅ Final expression: \( \boxed{a = \frac{\rho V g}{m} – g} \)

(c)
The expression from part B is consistent with the claim in part A.
Justification:
• In the expression \( a = \frac{\rho V g}{m} – g \), \( V, g, \) and \( m \) are constants.
• Acceleration \( a \) depends directly on fluid density \( \rho \) because \( \rho \) appears in the positive numerator term.
• Therefore, as \( \rho \) increases (salt water vs. freshwater), \( a \) increases.
• This matches the conclusion \( a_1 < a_2 \) since \( \rho_2 > \rho_1 \).
✅ Answer: Consistent, because \( a \) increases with \( \rho \).

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