Questions on Newton's laws with a variable force

Question 1

State the two forms of acceleration available, and the rule for choosing between them.

Accepted Answer

dv/dt and v dv/dx, which are equal by the chain rule. Use dv/dt when the force depends on time or on velocity and time is wanted; use v dv/dx when it depends on position, or on velocity and distance is wanted. The aim is an equation that separates into two variables. B1 for the two forms, B1 for the rule for choosing.

Question 2

A particle of mass 3 kg starts from rest at the origin under a force F = 12 − 3x newtons. Find its speed at x = 2.

Accepted Answer

The force depends on x, so use 3v dv/dx = 12 − 3x, that is v dv/dx = 4 − x. Integrating: ½v² = 4x − ½x² + c, and v = 0 at x = 0 gives c = 0. So v² = 8x − x², and at x = 2, v² = 16 − 4 = 12, giving v = 3.46 m/s. M1 for v dv/dx = 4 − x, A1 for ½v² = 4x − ½x² with c = 0, A1 for 3.46 m/s.

Question 3

A particle of mass 2 kg moves in a straight line with speed 10 m/s when it enters a region where the only force on it is a resistance of 4v newtons, v being its speed. Find the time for its speed to fall to 5 m/s.

Accepted Answer

The force depends on v and time is wanted, so 2 dv/dt = −4v, that is dv/dt = −2v. Separating: ln v = −2t + c, and v = 10 at t = 0 gives v = 10e^−2t. Setting v = 5: e^−2t = ½, so t = (ln 2)/2 = 0.347 s. M1 for 2 dv/dt = −4v, A1 for v = 10e^−2t, A1 for 0.347 s.

Question 4

A particle of mass 1.5 kg starts from rest at the origin and moves along the x-axis under a force of 3x² newtons in the positive direction. By finding the work done by the force, find the speed of the particle at x = 2.

Accepted Answer

Work done = ∫3x² dx from 0 to 2 = [x³] = 8 J. By the work-energy principle this is the kinetic energy gained: ½ × 1.5 × v² = 8, so v² = 32/3 and v = 3.27 m/s. M1 for integrating the force over the distance, A1 for 8 J, A1 for 3.27 m/s. Solving 1.5v dv/dx = 3x² gives 0.75v² = x³ and the same answer; the work integral is that equation integrated once already.

Question 5

A particle of mass 2 kg starts from rest at the origin under a force F = 6/(x + 1)² newtons. Find its speed at x = 2.

Accepted Answer

Again the force depends on x: 2v dv/dx = 6/(x + 1)². Integrating: v² = ∫6/(x + 1)² dx = −6/(x + 1) + c. With v = 0 at x = 0, c = 6, so v² = 6 − 6/(x + 1). At x = 2: v² = 6 − 2 = 4 and v = 2 m/s. M1 for 2v dv/dx = 6/(x + 1)², A1 for the integral, M1 for finding c, A1 for 2 m/s. However far it travels the speed cannot exceed √6, since the force dies away too fast.

Question 6

A planet has radius R and surface gravity g. Find the work done per unit mass against gravity in moving from the surface to a distance 3R from the centre.

Accepted Answer

At the surface the force per unit mass is g, so GM = gR² and at distance x it is gR²/x². Work = ∫gR²/x² dx from R to 3R = gR²[−1/x] = gR²(1/R − 1/(3R)) = 2gR/3. With Earth values that is 2(9.8)(6.4 × 10⁶)/3 ≈ 41.8 MJ per kilogram. B1 for GM = gR², M1 for integrating gR²/x² from R to 3R, A1 for 2gR/3, A1 for the numerical value. That is two thirds of the gR ≈ 62.7 MJ per kilogram needed to escape completely.

Question 7

Explain why the constant-acceleration equations cannot be used for a variable force, even with an average value of the force.

Accepted Answer

Those equations are derived by integrating a constant acceleration, so they are simply not solutions of the problem when a varies. Using the average force gives the right total impulse only if the force is averaged over time, and the right work only if it is averaged over distance, and those are different averages. Neither gives the correct answer for both. B1 for the derivation assuming constant acceleration, B1 for the time average giving impulse, B1 for the distance average giving work.

Question 8

A particle of mass 0.5 kg starts from rest and moves in a straight line under a force F = 6 − 2t newtons at time t seconds. Find its speed when t = 3, and verify your answer by using the impulse of the force.

Accepted Answer

The force depends on time, so use 0.5 dv/dt = 6 − 2t, that is dv/dt = 12 − 4t. Integrating: v = 12t − 2t² + c, and starting from rest gives c = 0. At t = 3, v = 36 − 18 = 18 m/s.
By impulse: ∫(6 − 2t) dt from 0 to 3 is [6t − t²] = 18 − 9 = 9 N s, and impulse equals change of momentum, so 0.5v = 9 and v = 18 m/s again. For a force given against time, the impulse route answers a speed-after-a-time question without solving the differential equation at all. M1 for dv/dt = 12 − 4t, A1 for v = 12t − 2t² with c = 0, A1 for 18 m/s, B1 for the impulse verification.

Question 9

A body is released from rest at a distance 2R from the centre of a planet of radius R and surface gravity g, and falls directly towards it under gravity alone. Show that it strikes the surface with speed √(gR), and evaluate this speed for R = 6400 km and g = 9.8 m/s². How does it compare with the escape speed?

Accepted Answer

At distance x from the centre the acceleration is −gR²/x², and distance is wanted, so use v dv/dx = −gR²/x². Integrating gives ½v² = gR²/x + c.
To escape, v must remain positive however large x becomes, and the limiting case has v tending to 0 as x grows without bound, so c = 0. At the surface ½v² = gR, giving v = √(2gR). Substituting, √(2 × 9.8 × 6.4 × 10⁶) = 11200 m/s, about 11.2 km/s.
For the second body, v² = 2gR/4 = gR/2 at x = R. Then ½(gR/2) = gR + c gives c = −3gR/4. The body stops where v = 0, that is gR²/x = 3gR/4, so x = 4R/3 from the centre and it rises R/3 above the surface, about 2130 km for the Earth.
M1 for v dv/dx = −gR²/x², A1 for ½v² = gR²/x + c, M1 for c = 0 in the limiting case, A1 for √(2gR), A1 for 11200 m/s, M1 for v² = gR/2 at the surface, A1 for the stopping distance from the centre, A1 for the rise above the surface.
The mass of the body plays no part in either result.

Question 10

A block of mass 3 kg rests on a rough horizontal floor, with coefficient of friction 0.2. It is pushed from rest by a horizontal force of (20 − 4x) newtons, where x metres is the distance it has moved. Find how far the block travels before it first comes to rest. Take g = 9.8 m/s².

Accepted Answer

Friction opposes the motion with 0.2 × 3 × 9.8 = 5.88 N, so the net force is 20 − 4x − 5.88 = 14.12 − 4x, which is positive at x = 0, so the block does start. With x wanted: 3v dv/dx = 14.12 − 4x. Integrating, 1.5v² = 14.12x − 2x², with c = 0 from rest at x = 0. The block stops where v = 0 again: x(14.12 − 2x) = 0, so x = 14.12/2 = 7.06 m. B1 for the friction 5.88 N in the equation, M1 for 3v dv/dx = 14.12 − 4x, A1 for 1.5v² = 14.12x − 2x², A1 for 7.06 m. The push itself becomes negative beyond x = 5, pulling back, yet the block keeps going for two more metres on the kinetic energy it has already gained; once it stops, the push of 20 − 28.24 = −8.24 N together with friction decides whether it stays put.

Question 11

A particle of mass 0.5 kg starts from rest at t = 0 and moves in a straight line under a force of 3 sin 2t newtons at time t seconds. Find its velocity at time t, its greatest speed, and the first time at which that greatest speed occurs.

Accepted Answer

0.5 dv/dt = 3 sin 2t, so dv/dt = 6 sin 2t and v = −3 cos 2t + c. From rest at t = 0, c = 3, so v = 3(1 − cos 2t). Since cos 2t ranges between −1 and 1, v ranges between 0 and 6: the greatest speed is 6 m/s, first reached when cos 2t = −1, that is 2t = π and t = π/2 ≈ 1.57 s. M1 for integrating dv/dt = 6 sin 2t, A1 for v = 3(1 − cos 2t), A1 for 6 m/s, A1 for t = π/2. The velocity is never negative, so the force, although it reverses every half-cycle, only ever slows the particle back to rest and never turns it round.

Question 12

Show that the escape speed from a planet of radius R and surface gravity g is √(2gR), and evaluate it for R = 6400 km and g = 9.8 m/s². A body is then projected vertically from the surface with half the escape speed. Find, in terms of R, how far above the surface it rises.

Accepted Answer

At distance x from the centre the acceleration is −gR²/x², and distance is wanted, so use v dv/dx = −gR²/x². Integrating gives ½v² = gR²/x + c.
To escape, v must remain positive however large x becomes, and the limiting case has v tending to 0 as x grows without bound, so c = 0. At the surface ½v² = gR, giving v = √(2gR). Substituting, √(2 × 9.8 × 6.4 × 10⁶) = 11200 m/s, about 11.2 km/s.
For the second body, v² = 2gR/4 = gR/2 at x = R. Then ½(gR/2) = gR + c gives c = −3gR/4. The body stops where v = 0, that is gR²/x = 3gR/4, so x = 4R/3 from the centre and it rises R/3 above the surface, about 2130 km for the Earth.
M1 for v dv/dx = −gR²/x², A1 for ½v² = gR²/x + c, M1 for c = 0 in the limiting case, A1 for √(2gR), A1 for 11200 m/s, M1 for v² = gR/2 at the surface, A1 for the stopping distance from the centre, A1 for the rise above the surface.
The mass of the body plays no part in either result.

Question 13

A particle of mass 2 kg starts from rest at time t = 0 and moves in a straight line under a force F = 24e^−2t newtons. Show that its speed never reaches 6 m/s, and find the time at which the speed is 3 m/s. Explain how the limiting speed could have been predicted from the total impulse of the force.

Accepted Answer

The force depends on time, so 2 dv/dt = 24e^−2t, that is dv/dt = 12e^−2t. Integrating: v = −6e^−2t + c, and v = 0 at t = 0 gives c = 6, so v = 6(1 − e^−2t).
Since e^−2t > 0 for every t, v < 6 always, though it approaches 6 m/s as t grows without bound.
Setting v = 3: e^−2t = ½, so 2t = ln 2 and t = (ln 2)/2 = 0.347 s. Half the limiting speed arrives in one half-life of the force.
The total impulse is ∫24e^−2t dt from 0 to ∞, which is 24/2 = 12 N s. Impulse equals change of momentum, so the final momentum is 12 N s and the limiting speed is 12/2 = 6 m/s, with no integration of the motion needed. M1 for dv/dt = 12e^−2t, A1 for v = 6(1 − e^−2t), B1 for v < 6 with the reason, M1 for setting v = 3, A1 for t = (ln 2)/2 ≈ 0.347 s, B1 for the impulse argument.

Question 14

A body is projected vertically upwards from the surface of a planet of radius R and surface gravity g with speed √(gR). Using the inverse square law, show that it rises to a height R above the surface. Show also that treating g as constant predicts a rise of only R/2, and explain the difference. Evaluate both heights for R = 6400 km.

Accepted Answer

With x measured from the centre, the acceleration is −gR²/x², so v dv/dx = −gR²/x² and ½v² = gR²/x + c. At x = R, v² = gR, so ½gR = gR + c and c = −gR/2. The body stops where gR²/x = gR/2, that is at x = 2R, a height R above the surface: 6400 km.
With g constant, v² = u² − 2gh gives h = gR/(2g) = R/2: 3200 km.
The constant-g model keeps the full surface weight acting all the way up, so it slows the body faster than gravity really does. Under the inverse square law the pull weakens as the body climbs, so the same launch energy carries it twice as far.
M1 for v dv/dx = −gR²/x², A1 for ½v² = gR²/x + c with c = −gR/2, A1 for x = 2R and the rise R, B1 for R/2 from the constant-g model, B1 for the explanation, A1 for 6400 km and 3200 km. The two models agree only for rises small compared with R, which is why constant g is adequate for a thrown ball and not for a launch to orbit.

Question 15

A particle of mass 2 kg is moving at 10 m/s when it enters a region where the only force on it is a resistance of (4 + 2v) newtons, v being its speed. Find the distance it travels and the time it takes before coming to rest.

Accepted Answer

Distance: 2v dv/dx = −(4 + 2v), so v dv/(2 + v) = −dx. Since v/(2 + v) = 1 − 2/(2 + v), integrating gives v − 2 ln(2 + v) = −x + c. At x = 0, v = 10: c = 10 − 2 ln 12. At rest v = 0: −2 ln 2 = −x + 10 − 2 ln 12, so x = 10 − 2 ln 12 + 2 ln 2 = 10 − 2 ln 6 = 6.42 m.
Time: 2 dv/dt = −(4 + 2v), so dv/(2 + v) = −dt and ln(2 + v) = −t + ln 12, that is 2 + v = 12e^−t. At rest 2 = 12e^−t, so t = ln 6 = 1.79 s.
M1 for 2v dv/dx = −(4 + 2v), M1 for splitting v/(2 + v) as 1 − 2/(2 + v), A1 for v − 2 ln(2 + v) = −x + c with c found, A1 for 6.42 m, M1 for the dv/dt equation separated, A1 for t = ln 6. The constant 4 N in the resistance is what brings the particle to rest in a finite time; a resistance of 2v alone would give the exponential v = 10e^−t and no stopping time at all.