University Physics I · unit UPI-08
Linear Momentum
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Linear Momentum & Systems: infer an input with uncertainty
Linear momentum is the vector p = mv, mass times velocity; a system's total momentum is the vector sum of the momenta of its parts, and only forces from outside the system can change it. Use the supplied topic model Y_out = ft, where Y_out is CART A dp_A/dt (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. ft is Winch tension F (internal), measured in N. Keep fe (Fan force E on cart B (external)) = 0 N; u (Observer velocity u) = 0 m s⁻¹ fixed. For this exercise use only 0.659375 N ≤ ft ≤ 1.84062 N on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.3799375 N, with a bounded reading uncertainty ±0.069103 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ft by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125 N, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CART A dp_A/dt against ft
Predictions from the supplied Linear Momentum & Systems model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CART A dp_A/dt / N
Horizontal axis (x): ft: Winch tension F (internal) / N
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ft: Winch tension F (internal) / N | CART A dp_A/dt / N |
|---|---|
| 0.659375 | 0.659375 |
| 0.9546875 | 0.9546875 |
| 1.25 | 1.25 |
| 1.5453125 | 1.5453125 |
| 1.840625 | 1.840625 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(ft) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.659375) = 0.659375 N and Y_out(1.84062) = 1.84062 N. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ft_line = 0.659375 + (1.37994 − 0.659375)(1.84062 − 0.659375)/(1.84062 − 0.659375) = 1.37994 N. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives ft = 1.37994 N; substituting back gives Y_out = 1.3799375 N.
- The interpolation error in the input is |ft_line − ft| = 0 N. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.3108345 N and 1.4490405 N. Sort the roots even if the response decreases: 1.31083 N ≤ ft ≤ 1.44904 N.
- The lower and upper input deviations are 0.069103 N and 0.069103 N. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer ft ≈ 1.37994 N; compatible interval [1.31083, 1.44904] N. Endpoint interpolation gives 1.37994 N.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Impulse and force–time graphs: infer an input with uncertainty
Impulse is the time integral of a force, J = ∫F dt, a vector measured in N s; the net impulse on a body equals its change in momentum, and for each component it is the signed area under the force–time graph. Use the supplied topic model Y_out = ct*acos((r-1)/(r+1))/pi, where Y_out is BALL AT REST AT t (ms). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. r is Rebound speed ÷ impact speed r, dimensionless. Keep v (Impact speed v (0.40 kg ball)) = 6 m s⁻¹; ct (Contact time τ) = 8 ms fixed. For this exercise use only 0.10625 ≤ r ≤ 0.89375 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 4.6711976 ms, with a bounded reading uncertainty ±0.079238 ms. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BALL AT REST AT t against r
Predictions from the supplied Impulse and force–time graphs model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BALL AT REST AT t / ms
Horizontal axis (x): r: Rebound speed ÷ impact speed r
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Rebound speed ÷ impact speed r | BALL AT REST AT t / ms |
|---|---|
| 0.10625 | 6.3952086242465125 |
| 0.303125 | 5.436820564230729 |
| 0.5 | 4.865387583755142 |
| 0.6968749999999999 | 4.457350696107122 |
| 0.8937499999999999 | 4.1429468267039296 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.10625) = 6.39521 ms and Y_out(0.89375) = 4.14295 ms. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.10625 + (4.6712 − 6.39521)(0.89375 − 0.10625)/(4.14295 − 6.39521) = 0.709048. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 0.586625; substituting back gives Y_out = 4.6711976 ms.
- The interpolation error in the input is |r_line − r| = 0.122423. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.5919596 ms and 4.7504356 ms. Sort the roots even if the response decreases: 0.549849 ≤ r ≤ 0.625522.
- The lower and upper input deviations are 0.0367761 and 0.038897. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 0.586625; compatible interval [0.549849, 0.625522]. Endpoint interpolation gives 0.709048.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Conservation of momentum: infer an input with uncertainty
A system's total momentum stays constant whenever the net external impulse on it is zero; forces between members of the system only move momentum from one member to another. Use the supplied topic model Y_out = 1200*(cosd(th)-mu*sind(th))/mc, where Y_out is RECOIL SPEED V (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. mc is Cannon mass M, measured in kg. Keep th (Launch angle θ) = 30 °; mu (Ground friction μ) = 0 fixed. For this exercise use only 242.5 kg ≤ mc ≤ 557.5 kg on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 2.3909594 m s⁻¹, with a bounded reading uncertainty ±0.07903 m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0315 kg, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RECOIL SPEED V against mc
Predictions from the supplied Conservation of momentum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RECOIL SPEED V / m s⁻¹
Horizontal axis (x): mc: Cannon mass M / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mc: Cannon mass M / kg | RECOIL SPEED V / m s⁻¹ |
|---|---|
| 242.5 | 4.285486534191037 |
| 321.25 | 3.2349587067434284 |
| 400 | 2.598076211353316 |
| 478.75 | 2.170716416796504 |
| 557.5 | 1.8640905552310787 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(mc) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(242.5) = 4.28549 m s⁻¹ and Y_out(557.5) = 1.86409 m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mc_line = 242.5 + (2.39096 − 4.28549)(557.5 − 242.5)/(1.86409 − 4.28549) = 488.96 kg. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives mc = 434.65 kg; substituting back gives Y_out = 2.3909594 m s⁻¹.
- The interpolation error in the input is |mc_line − mc| = 54.3095 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.3119294 m s⁻¹ and 2.4699894 m s⁻¹. Sort the roots even if the response decreases: 420.743 kg ≤ mc ≤ 449.508 kg.
- The lower and upper input deviations are 13.9071 kg and 14.8579 kg. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer mc ≈ 434.65 kg; compatible interval [420.743, 449.508] kg. Endpoint interpolation gives 488.96 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
One-Dimensional Collisions: infer an input with uncertainty
A one-dimensional collision is a brief contact between two bodies moving along one line: total momentum is unchanged across it, and the coefficient of restitution e, separation speed over approach speed, sets how that momentum is shared and how much kinetic energy survives. Use the supplied topic model Y_out = (100*(2+m2*v2+e*(2-v2))/(1+m2))/100, where Y_out is CART 2 AFTER v₂′ (before display rounding) (m/s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m2 is Mass of cart 2, m₂, measured in kg. Keep v2 (Initial velocity of cart 2, v₂) = 0 m/s; e (Coefficient of restitution e) = 0.5 fixed. For this exercise use only 0.648438 kg ≤ m2 ≤ 3.60156 kg on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.8696046 m/s, with a bounded reading uncertainty ±0.032648 m/s. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000295312 kg, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CART 2 AFTER v₂′ (before display rounding) against m2
Predictions from the supplied One-Dimensional Collisions model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CART 2 AFTER v₂′ (before display rounding) / m/s
Horizontal axis (x): m2: Mass of cart 2, m₂ / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m2: Mass of cart 2, m₂ / kg | CART 2 AFTER v₂′ (before display rounding) / m/s |
|---|---|
| 0.6484375 | 1.8199052132701423 |
| 1.38671875 | 1.2569558101472995 |
| 2.125 | 0.96 |
| 2.8632812499999996 | 0.776541961577351 |
| 3.6015624999999996 | 0.6519524617996605 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(m2) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.648438) = 1.81991 m/s and Y_out(3.60156) = 0.651952 m/s. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m2_line = 0.648438 + (0.869605 − 1.81991)(3.60156 − 0.648438)/(0.651952 − 1.81991) = 3.05124 kg. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives m2 = 2.44984 kg; substituting back gives Y_out = 0.8696046 m/s.
- The interpolation error in the input is |m2_line − m2| = 0.601393 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.8369566 m/s and 0.9022526 m/s. Sort the roots even if the response decreases: 2.32501 kg ≤ m2 ≤ 2.58442 kg.
- The lower and upper input deviations are 0.124833 kg and 0.134571 kg. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer m2 ≈ 2.44984 kg; compatible interval [2.32501, 2.58442] kg. Endpoint interpolation gives 3.05124 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.