University Physics I · Midterm Examination

A heavy, flexible rope of length L hangs freely from a hook. A small sideways flick at its lower end sends a pulse up the rope. A student proposes that the time t for the pulse to reach the hook depends only on L, the gravitational acceleration g (in m/s²) and the rope's mass per unit length μ (in kg/m), in the form t = C L^a g^b μ^c, where C is a dimensionless constant. Which exponents does dimensional analysis require?

1. Aa = ½, b = −½, c = 0
2. Ba = ½, b = −½, c = ½
3. Ca = −½, b = ½, c = 0
4. Da = ½, b = ½, c = 0

To find the average speed of a trolley, a student measures the distance between two marks on a bench as (1.250 ± 0.005) m and times the trolley between the marks as (0.50 ± 0.02) s. Which result is reported correctly?

1. A(2.500 ± 0.025) m/s
2. B(2.50 ± 0.25) m/s
3. C(2.50 ± 0.01) m/s
4. D(2.5 ± 0.1) m/s

A cart moves along a straight track with velocity v(t) = (6.0 − 2.0t) m/s, where t is in seconds, for 0 ≤ t ≤ 5.0 s. What distance does the cart travel in this interval?

1. A5.0 m
2. B9.0 m
3. C13 m
4. D30 m

A survey drone flies at 12.0 m/s relative to the air. A steady wind blows towards the east at 5.0 m/s relative to the ground. The pilot points the drone so that its path over the ground runs due north. What is the drone's speed relative to the ground?

1. A7.0 m/s
2. B10.9 m/s
3. C13.0 m/s
4. D17.0 m/s

A small magnet is held motionless on a vertical steel door. The door attracts it horizontally, and friction stops it sliding down. Which force is the Newton's-third-law partner of the upward friction force that the door exerts on the magnet?

1. AThe weight of the magnet: the downward gravitational force of the Earth on the magnet
2. BThe normal force that the door exerts on the magnet
3. CThe downward friction force that the magnet exerts on the door
4. DThe magnetic attraction that the door exerts on the magnet

A 62.0 kg passenger stands on bathroom scales in a lift. The lift is moving downwards and slowing down at 1.40 m/s². What do the scales read? Take g = 9.81 m/s².

1. A695 N
2. B608 N
3. C521 N
4. D86.8 N

A child on a playground swing is at the highest point of her swing, momentarily at rest, with the taut ropes at an angle to the vertical. Treat the child as a particle on a light rope of fixed length. What is the direction of her acceleration at that instant?

1. AShe has no acceleration, because her velocity is zero.
2. BVertically downwards, with magnitude g.
3. CAlong the rope, towards the pivot.
4. DPerpendicular to the rope, along the arc towards the lowest point.

A particle moves along the x-axis under a single conservative force. Its potential energy is U(x) = (3.0 J/m²)x² − (1.0 J/m³)x³, which has a minimum U = 0 at x = 0 and a maximum U = 4.0 J at x = 2.0 m. The particle is released from rest at x = −0.80 m. Which statement describes its subsequent motion?

1. AIt oscillates between x = −0.80 m and x = +0.80 m, because turning points lie equal distances either side of a minimum.
2. BIt oscillates between x = −0.80 m and a turning point that lies between x = +0.80 m and x = +2.0 m.
3. CIt passes x = +2.0 m and moves off towards large positive x.
4. DIt moves to x = 0, where U is least, and remains at rest there.

A campus fountain has its nozzle at ground level. The designer wants each jet of water to reach the highest point of its path exactly at the centre of a decorative ring, which is H = 1.80 m above the nozzle and d = 3.00 m from it horizontally. A low wall 0.60 m high stands 5.40 m horizontally from the nozzle, in the vertical plane of the jet, on the far side of the ring. Treat each parcel of water as a projectile launched from the nozzle at speed v₀ and angle θ above the horizontal, with no air resistance. Take g = 9.81 m/s².

1. (a)

   Show that Show that the launch angle must satisfy tan θ = 2H/d.

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   1. 1

      At the highest point the vertical velocity is zero, so the time to reach it is t₁ = v₀ sin θ/g and the height is H = (v₀ sin θ)²/(2g), equivalently H = ½gt₁²

   2. 2

      The horizontal velocity is constant, so d = (v₀ cos θ)t₁

   3. 3

      tan θ = (v₀ sin θ)/(v₀ cos θ) = gt₁/(d/t₁) = gt₁²/d = 2H/d

2. (b)

   Calculate Calculate the launch angle θ and the launch speed v₀.

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   1. 1

      tan θ = 2 × 1.80/3.00 = 1.20, so θ = 50.2°

   2. 2

      Vertical component v₀ sin θ = √(2gH) = √(2 × 9.81 × 1.80) = 5.94 m/s; time to the top t₁ = 5.94/9.81 = 0.606 s, so horizontal component v₀ cos θ = 3.00/0.606 = 4.95 m/s

   3. 3

      v₀ = √(5.94² + 4.95²) = 7.74 m/s

3. (c)

   Explain Explain why the launch angle in (a) does not depend on g, whereas the launch speed in (b) does.

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   1. 1

      The path of a projectile is a parabola, and a parabola that starts at the nozzle and has its vertex at the point (d, H) is fixed by those two points alone, so its initial slope tan θ = 2H/d is set by geometry

   2. 2

      g sets only how quickly the water travels along that parabola: a larger g needs larger velocity components for the same path (v₀ sin θ = √(2gH), so v₀ ∝ √g), leaving the direction unchanged

4. (d)

   Determine Determine the height of the jet as it passes the position of the wall, and state whether the jet clears the wall.

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   1. 1

      Using the symmetry of the parabola about x = d, y = H[1 − ((x − d)/d)²] = 1.80 × [1 − (2.40/3.00)²] = 0.648 m (or, with unrounded components, t = 5.40/4.952 = 1.090 s and y = 5.943t − 4.905t² = 0.648 m; accept 0.64 m to 0.65 m from rounded intermediate values)

   2. 2

      0.648 m is greater than 0.60 m, so the jet clears the wall, but only by about 0.05 m (4.8 cm)

5. (e)

   Predict In service, a partly blocked filter reduces the launch speed by 2.0 % while the launch angle stays the same. Predict whether the jet still clears the wall, supporting your prediction with a calculation.

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   1. 1

      The trajectory is y = x tan θ − gx²/(2v₀² cos²θ); at x = 5.40 m the first term is 6.48 m and the gravitational drop term is 5.83 m, which scales as 1/v₀² and so grows by a factor 1/0.98² = 1.041 to 6.07 m

   2. 2

      y = 6.48 − 6.07 = 0.41 m, below the 0.60 m wall, so the jet now strikes the wall: a 2 % loss of speed lowers the jet at the wall by about 0.24 m

Block A, of mass m_A = 4.00 kg, rests on a horizontal bench. A light inextensible cord is tied to A and runs horizontally to a light, frictionless pulley fixed at the edge of the bench. From that pulley the cord hangs vertically downwards, passes under a second light, frictionless pulley that is free to move, and rises vertically to a hook fixed to the underside of the bench top, so the two vertical sections of cord are parallel. Load B, of mass m_B = 6.00 kg, hangs from the axle of the movable pulley. The coefficients of friction between A and the bench are μ_s = 0.50 (static) and μ_k = 0.35 (kinetic). The system is released from rest. Take g = 9.81 m/s².

1. (a)

   Explain Explain why the magnitude of B's acceleration is half the magnitude of A's acceleration.

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   1. 1

      If A moves a distance s towards the edge, the horizontal section of cord shortens by s, so, because the cord is inextensible, the combined length of the two vertical sections increases by s

   2. 2

      That extra length is shared equally by the two vertical sections that support the movable pulley, so B descends s/2; differentiating twice with respect to time gives a_B = a_A/2

2. (b)

   Determine Determine the smallest mass of B that would set A in motion from rest, and hence confirm that A moves when the 6.00 kg load is used.

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   1. 1

      With the system at rest, the light movable pulley is in equilibrium under two upward tensions and B's weight, so 2T = m_B g

   2. 2

      A stays at rest only if the friction it needs, which equals T, does not exceed μ_s m_A g = 0.50 × 4.00 × 9.81 = 19.6 N

   3. 3

      Motion starts if m_B g/2 > μ_s m_A g, that is m_B > 2μ_s m_A = 4.00 kg; 6.00 kg exceeds this (it would need T = 29.4 N), so A moves

3. (c)

   Determine Determine the acceleration of A, the acceleration of B and the tension in the cord while the system is moving.

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   1. 1

      For A, horizontally: T − μ_k m_A g = m_A a_A, where μ_k m_A g = 0.35 × 4.00 × 9.81 = 13.7 N

   2. 2

      For B together with the movable pulley, vertically: m_B g − 2T = m_B a_B, with a_B = a_A/2

   3. 3

      Eliminating T: a_A = g(m_B − 2μ_k m_A)/(2m_A + m_B/2) = 9.81 × 3.20/11.00 = 2.85 m/s²

   4. 4

      a_B = a_A/2 = 1.43 m/s², downwards

   5. 5

      T = μ_k m_A g + m_A a_A = 13.7 N + 11.4 N = 25.1 N (check on B with unrounded values: 58.86 N − 2 × 25.15 N = 8.56 N, and 6.00 kg × 1.427 m/s² = 8.56 N)

4. (d)

   Evaluate A student treats A and B together as a single system and writes m_B g − μ_k m_A g = (m_A + m_B)a. Evaluate this equation as a model of the motion.

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   1. 1

      The equation gives both bodies one acceleration a, but a_B = a_A/2, and A accelerates horizontally while B accelerates vertically, so (m_A + m_B)a is not the rate of change of the system's momentum

   2. 2

      It treats the tension as internal and lets it cancel, but the hook and the fixed pulley lie outside the system and exert forces on the cord, so external forces are missing from the left-hand side (equally creditworthy: the cord pulls A with T but supports B with 2T, so the tension forces do not cancel)

   3. 3

      It predicts a = 9.81 × (6.00 − 1.40)/10.00 = 4.51 m/s², far above the correct 2.85 m/s², so the equation is not a valid model of this arrangement

A loaded truck of mass m = 1.80 × 10⁴ kg loses its brakes on a mountain road and enters an emergency arrester bed at v₀ = 25.0 m/s. The bed is a straight lane of loose gravel that rises uphill at a constant angle θ = 5.0° to the horizontal. The gravel gets deeper along the bed, and it exerts a resistive force on the truck, parallel to the slope and opposite to the motion, of magnitude F = μ(s)mg cos θ, where s is the distance travelled along the bed and μ(s) = μ₀ + βs, with μ₀ = 0.20 and β = 5.0 × 10⁻³ m⁻¹. Treat the truck as a particle and ignore air resistance. Take g = 9.81 m/s².

1. (a)

   Derive Starting from the definition of work, derive an expression for the work done on the truck by the gravel as the truck travels a distance s along the bed.

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   1. 1

      The force varies with position, so W = ∫₀ˢ F⋅ds′ with the force antiparallel to the displacement: W = −∫₀ˢ (μ₀ + βs′)mg cos θ ds′

   2. 2

      Integrating term by term: W = −mg cos θ[μ₀s′ + ½βs′²] evaluated from 0 to s

   3. 3

      W_gravel = −mg cos θ(μ₀s + ½βs²)

2. (b)

   Show that Show that the stopping distance s satisfies ½v₀² = gs sin θ + g cos θ(μ₀s + ½βs²).

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   1. 1

      Gravity does work −mgs sin θ as the truck rises a height s sin θ, and the normal force does no work because it is perpendicular to the displacement

   2. 2

      The work–kinetic energy theorem, W_net = 0 − ½mv₀², gives −mgs sin θ − mg cos θ(μ₀s + ½βs²) = −½mv₀², and dividing by m gives the stated result

3. (c)

   Calculate Calculate the stopping distance.

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   1. 1

      With g sin θ = 0.855 m/s² and g cos θ = 9.77 m/s², the equation becomes 0.0244s² + 2.81s − 312.5 = 0, with s in metres

   2. 2

      The physical, positive root is s = 69.4 m

4. (d)

   Determine Determine the percentage of the truck's initial kinetic energy that is stored as gravitational potential energy and the percentage that is dissipated in the gravel.

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   1. 1

      Gravitational potential energy gained = mgs sin θ = 1.80 × 10⁴ × 0.855 × 69.4 = 1.07 × 10⁶ J, which is 19.0 % of the initial kinetic energy ½mv₀² = 5.63 × 10⁶ J

   2. 2

      The remaining 4.56 × 10⁶ J, 81.0 %, is dissipated by the gravel's nonconservative force as thermal energy and deformation of the bed

5. (e)

   Evaluate A highway engineer argues that a heavier truck needs a longer arrester bed. Evaluate this argument, first within the model and then for a real bed.

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   1. 1

      Within the model both forces along the slope, the gravel's resistance and the weight component, are proportional to m, so m cancels from the work–energy equation and the stopping distance is independent of mass: the model does not support the argument

   2. 2

      A real bed has effects that do not scale with m — air resistance, set by the truck's size and speed, slows a heavier truck less and would lengthen its stop, while a heavier truck sinks deeper and meets a larger μ, which would shorten it — so the argument could hold only through such an unmodelled effect, and it must be tested rather than assumed

A cyclist rides a bicycle fitted with a power meter along a level road on a windless day. At each of six steady speeds v she records the power P that she delivers to the rear wheel. A model of the total resistance on the rider and bicycle is F = F_r + kv², where F_r is a constant rolling resistance and kv² is aerodynamic drag, with k constant. Take g = 9.81 m/s².

1. (a)

   Explain Explain why, at a steady speed on a level road, the model predicts that P = (F_r + kv²)v.

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   1. 1

      At constant velocity the acceleration is zero, so by Newton's second law the forward driving force equals the total resistive force F_r + kv²

   2. 2

      The power delivered by a force is P = F⋅v; the driving force is parallel to the velocity, so P = (F_r + kv²)v

2. (b)

   Determine By processing the data so that the model predicts a straight line, determine F_r and k.

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   1. 1

      Rearranges the model to P/v = F_r + kv², so a graph of P/v against v² should be a straight line with gradient k and intercept F_r

   2. 2

      Calculates P/v = 7.25, 11.3, 16.6, 24.3, 32.5 and 43.4 N for v² = 16, 36, 64, 100, 144 and 196 m²/s²

   3. 3

      Gradient of a best-fit line through all six points: k = 0.200 kg/m (accept 0.19 to 0.21 kg/m)

   4. 4

      Intercept: F_r = 4.0 N (accept 3.5 to 4.5 N)

3. (c)

   Compare A second student proposes that rolling resistance is negligible, so that P = kv³. Compare how well this model and the two-term model account for the data.

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   1. 1

      In the pure-drag model P/v³ should be constant, but it falls steadily from 0.453 kg/m at 4.0 m/s to 0.222 kg/m at 14.0 m/s: a systematic trend, not random scatter

   2. 2

      In the two-term model the values of P/v lie close to a straight line, with residuals within about ±0.35 N that are scattered in sign with no systematic trend, and the intercept of 4.0 N is clearly not zero — at 4.0 m/s it is more than half of the 7.25 N total resistance — so the rolling term is needed, above all at low speed

4. (d)

   Estimate Estimate the coefficient of rolling resistance C_r = F_r/N, where N is the normal force of the road on the bicycle and rider, stating the assumption you make.

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   1. 1

      States a plausible combined mass for rider and bicycle, for example 80 kg (any value from 60 kg to 100 kg), so that on a level road N = mg ≈ 80 × 9.81 = 785 N

   2. 2

      C_r = 4.0 N/785 N ≈ 0.005 (0.004 to 0.007 across the range of masses allowed)

5. (e)

   Describe Describe two changes to the procedure that would reduce the uncertainty in the value of F_r.

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   1. 1

      Add readings at lower speeds, for example 2.0 m/s to 4.0 m/s, where rolling resistance is the larger share of the total, so that the line is fixed close to its intercept rather than extrapolated to it

   2. 2

      Ride each speed on the same stretch of road in both directions, repeating and averaging, so that a slight gradient or breeze, which the model assumes absent, cancels instead of biasing the power readings

A snap-action switch is modelled as a slider of mass m that moves without friction along a straight horizontal rail, taken as the x-axis. A light spring of stiffness k and natural length L₀ joins the slider to a fixed pivot at perpendicular distance a from the rail, and the foot of that perpendicular is at x = 0. The pivot is closer to the rail than the spring's natural length, so a < L₀. Take the elastic potential energy of the spring to be zero at its natural length. Gravity and the normal force of the rail are perpendicular to the rail and do no work on the slider.

1. (a)

   Derive Starting from the elastic potential energy of a spring, ½k × (extension)², derive an expression for the potential energy U(x) of the system when the slider is at position x.

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   1. 1

      The spring runs from the pivot to the slider, so its length is √(x² + a²) and its extension is √(x² + a²) − L₀ (negative when the spring is compressed)

   2. 2

      Elastic potential energy ½k(extension)² gives U(x) = ½k(√(x² + a²) − L₀)²

2. (b)

   Derive Using the relation between a conservative force and its potential-energy function, derive an expression for the component F_x(x) of the force on the slider along the rail.

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   1. 1

      For a conservative force in one dimension, F_x = −dU/dx

   2. 2

      By the chain rule, dU/dx = k(√(x² + a²) − L₀) × x/√(x² + a²)

   3. 3

      F_x = −kx(1 − L₀/√(x² + a²))

3. (c)

   Determine Determine all the equilibrium positions of the slider.

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   1. 1

      F_x = 0 requires either x = 0 or √(x² + a²) = L₀

   2. 2

      The equilibria are x = 0 and x = ±√(L₀² − a²), the outer pair existing because a < L₀

4. (d)

   Justify Justify, for each equilibrium position, whether it is stable or unstable.

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   1. 1

      At x = ±√(L₀² − a²), U = 0, which is the least value U can take because it is ½k times a square, so these are minima of U and the equilibria are stable

   2. 2

      At x = 0, U = ½k(L₀ − a)² > 0 is a maximum between the two minima; equivalently, for small x, F_x ≈ kx(L₀/a − 1) has the same sign as x and pushes the slider away, so x = 0 is unstable

5. (e)

   Sketch Sketch U(x) against x over a range that includes all the equilibrium positions, marking each position and the value of U there.

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   1. 1

      A curve symmetric about x = 0 with two minima at x = ±√(L₀² − a²), where it touches U = 0 without crossing below the axis

   2. 2

      A smooth local maximum (zero slope, no cusp) at x = 0, marked at height ½k(L₀ − a)²

   3. 3

      U rising without limit on both sides beyond the minima, growing roughly as ½kx² once |x| is much larger than L₀, so that the whole curve is a double well that never falls below zero

6. (f)

   Calculate For a switch with k = 250 N/m, L₀ = 50.0 mm, a = 40.0 mm and m = 20.0 g, calculate the positions of the two stable states and the minimum speed the slider must have as it passes through one stable state if it is to reach the other.

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   1. 1

      Stable states at x = ±√(0.0500² − 0.0400²) m = ±0.0300 m, that is ±30.0 mm

   2. 2

      Height of the energy barrier: U(0) − 0 = ½ × 250 × (0.0500 − 0.0400)² = 1.25 × 10⁻² J

   3. 3

      Conservation of mechanical energy, ½mv² = 1.25 × 10⁻² J, gives v = √(2 × 1.25 × 10⁻²/0.0200) = 1.12 m/s (strictly, any speed greater than this)

7. (g)

   Predict The pivot is moved further from the rail until a ≥ L₀. Predict how the switch now behaves, reasoning from the energy landscape.

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   1. 1

      For a ≥ L₀ the spring's length √(x² + a²) is never less than a, so the spring is never compressed; √(x² + a²) = L₀ has no solution when a > L₀ and only x = 0 when a = L₀, so x = 0 is the only equilibrium: as a rises towards L₀ the outer minima at ±√(L₀² − a²) move in to x = 0 and the barrier ½k(L₀ − a)² shrinks to zero

   2. 2

      U now rises steadily with |x| from a single minimum at x = 0 (a very flat one when a = L₀), so x = 0 is stable: a displaced slider is always pushed back towards the centre and oscillates about it, so the switch loses its two latched states and no longer snaps