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Newton's laws with a variable force questions
When the force depends on where the particle is, F = ma becomes a differential equation. Which form of the acceleration to use is decided entirely by what the force depends on.
6 original questions · 24 marks · the newton's laws with a variable force notes · Further Mechanics 2
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State the two forms of acceleration available, and the rule for choosing between them.
Worked 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.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.
Worked 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.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.
Worked 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.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.
Worked 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.Explain why the constant-acceleration equations cannot be used for a variable force, even with an average value of the force.
Worked 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.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.
Worked 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.
Half the escape speed carries a quarter of the energy and buys only a third of a radius. The mass of the body plays no part in either result.
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