AS & A Level Further Mathematics (9231)
Linear motion under a variable force Banker
9 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Force depending on velocity · Core · Paper 3 style · 8 marksA particle P of mass 2 kg moves in a straight line on a smooth horizontal surface. The only horizontal force on P is a resistance of magnitude 4v N, where v m s⁻¹ is its speed. Initially P has speed 10 m s⁻¹.
- (a) Show that v = 10e⁻²ᵗ, where v m s⁻¹ is the speed of P at time t s. [3]
- (b) Find the time taken for the speed of P to fall to 2 m s⁻¹. [2]
- (c) Find the distance travelled by P in this time. [3]
Show answer
- (a) Shown
M1 Newton's second law: 2 dv/dt = −4v; M1 separate and integrate: ln v = −2t + c; A1 v = 10 at t = 0 gives v = 10e⁻²ᵗ.
- (b) 0.805 s
M1 10e⁻²ᵗ = 2; A1 t = (1/2) ln 5 = 0.805 s.
- (c) 4 m
M1 x = ∫ 10e⁻²ᵗ dt = 5(1 − e⁻²ᵗ); M1 e⁻²ᵗ = 1/5; A1 x = 5(4/5) = 4 m.
An ice-hockey puck of mass 0.5 kg slides in a straight line across horizontal ice. Air resistance and friction together are modelled as a single resisting force of magnitude 0.1v² N, where v m s⁻¹ is the speed. The puck starts with speed 20 m s⁻¹.
- (a) Show that v = 20e^(−0.2x), where v m s⁻¹ is the speed of the puck when it has travelled x m. [4]
- (b) Find the distance travelled when the speed has fallen to 5 m s⁻¹. [2]
Show answer
- (a) Shown
M1 use acceleration v dv/dx: 0.5v dv/dx = −0.1v²; A1 dv/dx = −0.2v; M1 integrate: ln v = −0.2x + c; A1 v = 20 at x = 0 gives result.
- (b) 6.93 m
M1 e^(−0.2x) = 1/4; A1 x = 5 ln 4 = 6.93 m.
A car of mass 800 kg starts from rest on a straight horizontal road. The engine produces a constant driving force of 2400 N and the resistance to motion is 40v N, where v m s⁻¹ is the speed of the car after t seconds.
- (a) Find the greatest speed the car could approach under this model. [1]
- (b) Show that v = 60(1 − e^(−t/20)). [4]
- (c) Find the time taken for the car to reach 30 m s⁻¹. [2]
Show answer
- (a) 60 m s⁻¹
B1 2400 − 40v = 0 when v = 60.
- (b) Shown
M1 800 dv/dt = 2400 − 40v; M1 separate: ∫ dv/(60 − v) = ∫ dt/20; A1 −ln(60 − v) = t/20 + c; A1 v = 0 at t = 0 gives result.
- (c) 13.9 s
M1 e^(−t/20) = 1/2; A1 t = 20 ln 2 = 13.9 s.
A small stone of mass 0.2 kg is released from rest and falls vertically through the air. The air resistance is 0.05v N, where v m s⁻¹ is the speed after t s. Take g = 10 m s⁻².
- (a) Show that dv/dt = 10 − 0.25v and hence find v in terms of t. [4]
- (b) Find the speed of the stone after 3 seconds. [1]
- (c) State the terminal speed of the stone. [1]
Show answer
- (a) v = 40(1 − e^(−t/4))
M1 0.2 dv/dt = 0.2(10) − 0.05v; A1 dv/dt = 10 − 0.25v; M1 separate and integrate: −4 ln(40 − v) = t + c; A1 v = 40(1 − e^(−t/4)).
- (b) 21.1 m s⁻¹
B1 40(1 − e^(−0.75)) = 21.1 m s⁻¹.
- (c) 40 m s⁻¹
B1 v → 40 as t → ∞.
A cyclist and her bicycle have total mass 80 kg. She rides along a straight horizontal road in Islamabad, working at a constant power of 240 W. The resistance to motion is 2v N, where v m s⁻¹ is her speed.
- (a) Show that 80v dv/dt = 240 − 2v². [2]
- (b) Find the greatest speed that the cyclist can approach. [2]
- (c) Find the acceleration of the cyclist when her speed is 5 m s⁻¹. [2]
Show answer
- (a) Shown
M1 driving force = power/speed = 240/v; A1 80 dv/dt = 240/v − 2v, multiply by v.
- (b) √120 ≈ 11.0 m s⁻¹
M1 240 − 2v² = 0; A1 v = √120 = 11.0 m s⁻¹.
- (c) 0.475 m s⁻²
M1 80a = 240/5 − 2(5); A1 a = 38/80 = 0.475 m s⁻².
A particle P of mass 3 kg moves along the positive x-axis on a smooth horizontal surface. It passes through the origin O with speed 8 m s⁻¹ and is subject to a resisting force of magnitude 6v N, where v m s⁻¹ is its speed when its displacement from O is x m.
- (a) Using acceleration in the form v dv/dx, show that v = 8 − 2x. [3]
- (b) Find the value of x when v = 2, and explain why P never reaches the point where x = 4. [3]
- (c) Find the time taken for P to travel from O to the point where x = 3. [2]
Show answer
- (a) Shown
M1 3v dv/dx = −6v; A1 dv/dx = −2; A1 v = 8 − 2x using v = 8 at x = 0.
- (b) x = 3; v → 0 only as t → ∞
B1 x = 3; M1 from 3 dv/dt = −6v, v = 8e⁻²ᵗ, which is never zero; A1 so x = 4 − 4e⁻²ᵗ < 4 for all t.
- (c) 0.693 s
M1 8e⁻²ᵗ = 2; A1 t = (1/2) ln 4 = 0.693 s.
A ball of mass 0.1 kg is projected vertically upwards with speed 15 m s⁻¹. Air resistance has magnitude 0.02v² N, where v m s⁻¹ is the speed of the ball. Take g = 10 m s⁻².
- (a) Find the greatest height above the point of projection reached by the ball. [4]
- (b) Find the time taken to reach this greatest height. [4]
Show answer
- (a) 4.26 m
M1 0.1v dv/dx = −1 − 0.02v²; A1 v dv/dx = −(10 + 0.2v²); M1 x = ∫₀¹⁵ v/(10 + 0.2v²) dv = 2.5 ln((10 + 45)/10); A1 2.5 ln 5.5 = 4.26 m.
- (b) 0.799 s
M1 dv/dt = −0.2(50 + v²); M1 t = 5∫₀¹⁵ dv/(50 + v²); A1 = (5/√50) tan⁻¹(15/√50); A1 0.799 s.
A car of mass 1200 kg moves along a straight horizontal motorway between Lahore and Islamabad. Its engine works at a constant rate of 30 kW and the resistance to motion is 25v N, where v m s⁻¹ is the speed.
- (a) Find the greatest possible speed of the car under this model. [2]
- (b) Show that 1200v dv/dt = 30 000 − 25v². [1]
- (c) Find the time taken for the speed to increase from 10 m s⁻¹ to 20 m s⁻¹. [4]
Show answer
- (a) √1200 ≈ 34.6 m s⁻¹
M1 30000/v = 25v; A1 v = √1200 = 34.6 m s⁻¹.
- (b) Shown
B1 1200 dv/dt = 30 000/v − 25v, multiplied by v.
- (c) 7.64 s
M1 separate: t = ∫ 1200v/(30 000 − 25v²) dv; M1 = −24 ln(30 000 − 25v²); A1 24 ln(27 500/20 000); A1 7.64 s.
A block of mass 2 kg slides on a horizontal surface. The total resistance to its motion is (4 + 2v) N, where v m s⁻¹ is its speed. The block is set moving with speed 6 m s⁻¹.
- (a) Find the time taken for the block to come to rest. [3]
- (b) Find the distance travelled by the block before it comes to rest. [4]
Show answer
- (a) ln 4 ≈ 1.39 s
M1 2 dv/dt = −(4 + 2v); M1 ∫₆⁰ dv/(2 + v) = −t; A1 t = ln 4 = 1.39 s.
- (b) 6 − 2 ln 4 ≈ 3.23 m
M1 v dv/dx = −(2 + v); M1 x = ∫₀⁶ v/(2 + v) dv = ∫₀⁶ (1 − 2/(2 + v)) dv; A1 6 − 2 ln 4; A1 3.23 m.
A particle of mass 0.4 kg is released from rest at a great height and falls vertically. Air resistance has magnitude 0.004v² N, where v m s⁻¹ is the speed. Take g = 10 m s⁻².
- (a) Find the terminal speed of the particle. [2]
- (b) Find the distance fallen when the speed is 20 m s⁻¹. [4]
Show answer
- (a) √1000 ≈ 31.6 m s⁻¹
M1 4 = 0.004v²; A1 v = √1000 = 31.6 m s⁻¹.
- (b) 50 ln(5/3) ≈ 25.5 m
M1 v dv/dx = 10 − 0.01v²; M1 x = ∫₀²⁰ v/(10 − 0.01v²) dv = −50 ln(10 − 0.01v²); A1 50 ln(10/6); A1 25.5 m.
A particle of mass 5 kg is projected with speed 12 m s⁻¹ up a line of greatest slope of a smooth plane inclined at 30° to the horizontal. Air resistance has magnitude 0.5v² N, where v m s⁻¹ is the speed. Take g = 10 m s⁻².
- (a) Show that, while the particle moves up the plane, v dv/dx = −(5 + 0.1v²), where x m is the distance travelled up the plane. [2]
- (b) Find the distance the particle travels up the plane before coming to instantaneous rest. [4]
Show answer
- (a) Shown
M1 5v dv/dx = −5g sin 30° − 0.5v²; A1 divide by 5: v dv/dx = −(5 + 0.1v²).
- (b) 5 ln 3.88 ≈ 6.78 m
M1 x = ∫₀¹² v/(5 + 0.1v²) dv; M1 = 5 ln((5 + 14.4)/5); A1 5 ln 3.88; A1 6.78 m.
A motor boat of mass 400 kg is moving in a straight line across a lake at 9 m s⁻¹ when its engine is switched off. The water resistance is modelled as 50v^(3/2) N, where v m s⁻¹ is the speed.
- (a) Find the time taken for the speed to fall from 9 m s⁻¹ to 4 m s⁻¹. [4]
- (b) Find the distance travelled in this time. [3]
Show answer
- (a) 8/3 s ≈ 2.67 s
M1 400 dv/dt = −50v^(3/2); M1 ∫ v^(−3/2) dv = −t/8; A1 2(1/√4 − 1/√9) = t/8; A1 t = 8/3 s.
- (b) 16 m
M1 400v dv/dx = −50v^(3/2) so dv/dx = −√v/8; M1 2(√9 − √4) = x/8; A1 x = 16 m.
A train of mass 50 000 kg starts from rest on a straight level track. The engine exerts a constant driving force of 40 000 N and the resistance to motion is kv N, where v m s⁻¹ is the speed and k is a constant.
- (a) Given that the greatest speed the train can approach is 40 m s⁻¹, show that k = 1000. [1]
- (b) Find the time taken for the train to reach 20 m s⁻¹ from rest. [3]
- (c) Find the distance travelled in this time. [4]
Show answer
- (a) Shown
B1 at top speed 40 000 = 40k so k = 1000.
- (b) 50 ln 2 ≈ 34.7 s
M1 50 000 dv/dt = 40 000 − 1000v; M1 t = 50∫₀²⁰ dv/(40 − v); A1 50 ln 2 = 34.7 s.
- (c) 386 m
M1 50v dv/dx = 40 − v; M1 x = 50∫₀²⁰ (40/(40 − v) − 1) dv; A1 50(40 ln 2 − 20); A1 386 m.
A particle of mass m kg is projected vertically upwards with speed 30 m s⁻¹. Air resistance has magnitude 0.01mv² N, where v m s⁻¹ is the speed, and acts opposite to the direction of motion. Take g = 10 m s⁻².
- (a) Find the greatest height reached above the point of projection. [4]
- (b) Find the speed of the particle when it returns to the point of projection. [5]
Show answer
- (a) 50 ln 1.9 ≈ 32.1 m
M1 upwards: v dv/dx = −(10 + 0.01v²); M1 H = ∫₀³⁰ v/(10 + 0.01v²) dv = 50 ln((10 + 9)/10); A1 50 ln 1.9; A1 32.1 m.
- (b) 21.8 m s⁻¹
M1 downwards: v dv/dy = 10 − 0.01v², y measured down from the top; M1 H = −50 ln(1 − 0.001w²); M1 equate: 1/(1 − 0.001w²) = 1.9; A1 w² = 1000(1 − 1/1.9) = 473.7; A1 w = 21.8 m s⁻¹.
A racing cyclist and her bicycle have total mass 75 kg. She works at a constant power of 300 W on a straight horizontal track, and the air resistance is 3v² N, where v m s⁻¹ is her speed when she has travelled x m.
- (a) Show that 75v² dv/dx = 300 − 3v³. [3]
- (b) Find the greatest speed the cyclist can approach, correct to 3 significant figures. [1]
- (c) Find the distance travelled while the speed increases from 2 m s⁻¹ to 4 m s⁻¹. [4]
Show answer
- (a) Shown
M1 driving force 300/v; M1 75v dv/dx = 300/v − 3v²; A1 multiply by v.
- (b) ∛100 ≈ 4.64 m s⁻¹
B1 3v³ = 300, v = ∛100 = 4.64 m s⁻¹.
- (c) (25/3) ln(23/9) ≈ 7.82 m
M1 x = ∫₂⁴ 75v²/(300 − 3v³) dv; M1 = −(25/3) ln(300 − 3v³); A1 (25/3) ln(276/108); A1 7.82 m.
A sledge of mass 0.8 kg slides across smooth horizontal ice in a straight line. It passes the point O with speed 20 m s⁻¹, and air resistance has magnitude 0.04v² N, where v m s⁻¹ is the speed.
- (a) Show that v = 20/(1 + t), where t s is the time after the sledge leaves O. [3]
- (b) Find x in terms of t, where x m is the distance from O. [2]
- (c) Find the time taken to travel 30 m from O, and the speed at that point. [3]
Show answer
- (a) Shown
M1 0.8 dv/dt = −0.04v²; M1 ∫ v⁻² dv = −0.05t gives 1/v = 0.05t + 1/20; A1 v = 20/(1 + t).
- (b) x = 20 ln(1 + t)
M1 x = ∫ 20/(1 + t) dt; A1 20 ln(1 + t).
- (c) t = e^1.5 − 1 ≈ 3.48 s; v = 20e^(−1.5) ≈ 4.46 m s⁻¹
M1 ln(1 + t) = 1.5; A1 t = 3.48 s; B1 v = 20/e^1.5 = 4.46 m s⁻¹.
A package of mass 1 kg is dropped from rest from a hovering drone. Air resistance has magnitude 0.5v N, where v m s⁻¹ is the speed of the package t s after release. Take g = 10 m s⁻².
- (a) Show that v = 20(1 − e^(−t/2)) and that the distance fallen is x = 20t − 40(1 − e^(−t/2)). [5]
- (b) The package is released 50 m above the ground. Show that the time T s taken to reach the ground satisfies T = 4.5 − 2e^(−T/2), and use an iterative method starting with T₀ = 4.5 to find T correct to 2 decimal places. [4]
Show answer
- (a) Shown
M1 dv/dt = 10 − 0.5v; A1 v = 20(1 − e^(−t/2)); M1 x = ∫ v dt; A1 20t + 40e^(−t/2) + c; A1 c = −40 from x = 0 at t = 0.
- (b) T ≈ 4.26
M1 20T − 40 + 40e^(−T/2) = 50; A1 divide by 20 and rearrange; M1 iterate Tₙ₊₁ = 4.5 − 2e^(−Tₙ/2): T₁ = 4.2892, T₂ = 4.2658, T₃ = 4.2630, T₄ = 4.2627, …; A1 T = 4.26.
A puck of mass m kg slides in a straight line on a rough horizontal floor, with coefficient of friction 0.2. In addition, air resistance of magnitude 0.1mv² N acts, where v m s⁻¹ is the speed. The initial speed is 10 m s⁻¹. Take g = 10 m s⁻².
- (a) Show that the deceleration of the puck is (2 + 0.1v²) m s⁻². [2]
- (b) Find the time taken for the puck to come to rest. [4]
- (c) Find the distance travelled before it comes to rest. [3]
Show answer
- (a) Shown
M1 friction μmg = 0.2 × 10 × m = 2m N; A1 total resisting force m(2 + 0.1v²), divide by m.
- (b) √5 tan⁻¹(√5) ≈ 2.57 s
M1 t = ∫₀¹⁰ dv/(2 + 0.1v²) = 10∫₀¹⁰ dv/(20 + v²); M1 = (10/√20) tan⁻¹(v/√20); A1 √5 tan⁻¹ √5; A1 2.57 s.
- (c) 5 ln 6 ≈ 8.96 m
M1 x = ∫₀¹⁰ v/(2 + 0.1v²) dv; M1 5 ln((2 + 10)/2); A1 5 ln 6 = 8.96 m.
A stone of mass 0.5 kg is thrown vertically upwards with speed 20 m s⁻¹. Air resistance has magnitude 0.25v N, where v m s⁻¹ is the speed. Take g = 10 m s⁻².
- (a) Find the time taken for the stone to reach its highest point. [3]
- (b) Find the greatest height reached, giving your answer in exact form and correct to 3 significant figures. [4]
Show answer
- (a) 2 ln 2 ≈ 1.39 s
M1 0.5 dv/dt = −5 − 0.25v so dv/dt = −(10 + 0.5v); M1 t = ∫₀²⁰ dv/(10 + 0.5v) = 2 ln 2; A1 1.39 s.
- (b) 40 − 40 ln 2 ≈ 12.3 m
M1 v dv/dx = −(10 + 0.5v); M1 H = ∫₀²⁰ v/(10 + 0.5v) dv = ∫₀²⁰ (2 − 20/(10 + 0.5v)) dv; A1 40 − 40 ln 2; A1 12.3 m.
A particle P of mass 0.5 kg starts from rest at a point O and moves in a straight line on a smooth horizontal surface. At time t s it is acted on by a single horizontal force of magnitude 4e⁻ᵗ N in the direction of motion. Its speed is v m s⁻¹ and its displacement from O is x m.
- (a) Find v in terms of t, and state the limiting value of v. [3]
- (b) Find the speed and the displacement of P from O when t = 2. [4]
Show answer
- (a) v = 8(1 − e⁻ᵗ); limit 8 m s⁻¹
M1 0.5 dv/dt = 4e⁻ᵗ; A1 v = −8e⁻ᵗ + c with v = 0 at t = 0, v = 8(1 − e⁻ᵗ); B1 v → 8.
- (b) v ≈ 6.92 m s⁻¹; x = 8 + 8e⁻² ≈ 9.08 m
B1 v = 8(1 − e⁻²) = 6.92; M1 x = ∫ 8(1 − e⁻ᵗ) dt = 8t + 8e⁻ᵗ + c; M1 c = −8; A1 x(2) = 8 + 8e⁻² = 9.08 m.
A particle of mass 3 kg moves in a straight line on a smooth horizontal surface under a single force of magnitude 12/(t + 1)² N, acting in the direction of motion, where t s is the time. At t = 0 its speed is 2 m s⁻¹.
- (a) Show that v = 6 − 4/(t + 1). [3]
- (b) Find the time at which the speed is 5 m s⁻¹, and the distance travelled up to that time. [4]
Show answer
- (a) Shown
M1 3 dv/dt = 12/(t + 1)²; A1 v = −4/(t + 1) + c; A1 v = 2 at t = 0 gives c = 6.
- (b) t = 3 s; x = 18 − 4 ln 4 ≈ 12.5 m
B1 4/(t + 1) = 1 so t = 3; M1 x = ∫₀³ (6 − 4/(t + 1)) dt; A1 18 − 4 ln 4; A1 12.5 m.
A particle of mass 1 kg starts from rest and moves in a straight line on a horizontal surface. At time t s it is acted on by a force of magnitude 4t N in the direction of motion and a resistance of magnitude 2v N, where v m s⁻¹ is its speed.
- (a) Show that dv/dt + 2v = 4t and hence find v in terms of t. [5]
- (b) Find the distance travelled in the first 2 seconds. [3]
Show answer
- (a) v = 2t − 1 + e⁻²ᵗ
B1 1 × dv/dt = 4t − 2v; M1 integrating factor e²ᵗ; M1 ve²ᵗ = ∫ 4te²ᵗ dt = (2t − 1)e²ᵗ + c; A1 c = 1 from v(0) = 0; A1 v = 2t − 1 + e⁻²ᵗ.
- (b) 2.5 − 0.5e⁻⁴ ≈ 2.49 m
M1 v > 0 for t > 0, so distance = ∫₀² (2t − 1 + e⁻²ᵗ) dt; M1 [t² − t − e⁻²ᵗ/2]₀²; A1 2.5 − 0.5e⁻⁴ = 2.49 m.
A particle of mass 2 kg starts from rest at O and moves along a straight line on a smooth horizontal surface under a force of 6 sin 3t N, where t s is the time, acting along the line. Its velocity is v m s⁻¹.
- (a) Find v in terms of t. [3]
- (b) Find the greatest speed of the particle and the first time at which it occurs. [2]
- (c) Find the displacement of the particle from O at this time. [3]
Show answer
- (a) v = 1 − cos 3t
M1 2 dv/dt = 6 sin 3t; A1 v = −cos 3t + c; A1 c = 1.
- (b) 2 m s⁻¹ at t = π/3
B1 greatest value of 1 − cos 3t is 2; B1 when 3t = π, t = π/3.
- (c) π/3 ≈ 1.05 m
M1 x = ∫ (1 − cos 3t) dt = t − (1/3) sin 3t; A1 c = 0; A1 x = π/3 = 1.05 m.
A particle of mass 2 kg moves in a straight line on a horizontal surface. At time t s it is acted on by a force of 4t N in the direction of motion and a resistance of 4v N, where v m s⁻¹ is its speed. When t = 0, v = 3.
- (a) Show that v = t − 1/2 + (7/2)e⁻²ᵗ. [5]
- (b) Find the least speed of the particle, giving your answer in exact form. [4]
Show answer
- (a) Shown
M1 2 dv/dt = 4t − 4v so dv/dt + 2v = 2t; M1 integrating factor e²ᵗ; M1 ve²ᵗ = ∫ 2te²ᵗ dt = (t − 1/2)e²ᵗ + c; A1 c = 7/2 from v(0) = 3; A1 result.
- (b) (1/2) ln 7 ≈ 0.973 m s⁻¹
M1 dv/dt = 1 − 7e⁻²ᵗ = 0; A1 t = (1/2) ln 7; M1 v = (1/2) ln 7 − 1/2 + (7/2)(1/7); A1 (1/2) ln 7 = 0.973 m s⁻¹.
A small robot of mass 5 kg starts from rest and moves along a straight horizontal track in a warehouse in Karachi. At time t s the resultant force on it is 10/(1 + t) N in the direction of motion. Its speed is v m s⁻¹.
- (a) Find v in terms of t. [2]
- (b) Find the time at which v = 4 and the distance travelled up to that time, giving the distance in exact form. [5]
Show answer
- (a) v = 2 ln(1 + t)
M1 5 dv/dt = 10/(1 + t); A1 v = 2 ln(1 + t) using v(0) = 0.
- (b) t = e² − 1 ≈ 6.39 s; x = 2(e² + 1) ≈ 16.8 m
B1 t = e² − 1; M1 x = ∫ 2 ln(1 + t) dt by parts; A1 x = 2[(1 + t) ln(1 + t) − t]; M1 substitute t = e² − 1: 2[2e² − e² + 1]; A1 2(e² + 1) = 16.8 m.
A particle P of mass 0.5 kg moves along the x-axis. It starts from rest at the origin O and is acted on by a force of (4 − 2x) N in the positive x-direction, where x m is its displacement from O. Its speed is v m s⁻¹.
- (a) Show that v² = 16x − 4x². [3]
- (b) Find the greatest speed of P and the value of x at which P next comes to rest. [3]
Show answer
- (a) Shown
M1 0.5v dv/dx = 4 − 2x; M1 integrate: v²/4 = 4x − x² + c; A1 c = 0, v² = 16x − 4x².
- (b) 4 m s⁻¹; x = 4
M1 force zero (or v² greatest) at x = 2; A1 v² = 32 − 16 = 16, v = 4; B1 v = 0 when x = 4.
A particle of mass 1.5 kg starts from rest at a point O and moves along a straight line. When its displacement from O is x m, the only force acting on it is 3/(x + 1)² N directed away from O. Its speed is v m s⁻¹.
- (a) Show that v² = 4x/(x + 1). [4]
- (b) Find the speed of the particle when x = 3, and state the value that the speed approaches as x increases. [2]
Show answer
- (a) Shown
M1 1.5v dv/dx = 3/(x + 1)²; M1 v²/2 = −2/(x + 1) + c; A1 c = 2; A1 v² = 4 − 4/(x + 1) = 4x/(x + 1).
- (b) √3 ≈ 1.73 m s⁻¹; approaches 2 m s⁻¹
B1 v² = 12/4 = 3; B1 v → 2.
A particle of mass 1 kg moves along the positive x-axis. It is projected from the origin O with speed 5 m s⁻¹ in the positive x-direction. When its displacement from O is x m, it is acted on by a single force of magnitude (2x + 3x²) N directed towards O. Its speed is v m s⁻¹.
- (a) Show that v² = 25 − 2x² − 2x³. [3]
- (b) Find the distance from O at which the particle first comes to instantaneous rest, giving your answer correct to 3 significant figures. [3]
Show answer
- (a) Shown
M1 v dv/dx = −(2x + 3x²); M1 v²/2 = −x² − x³ + c; A1 c = 12.5.
- (b) 2.03 m
M1 x³ + x² = 12.5; M1 numerical method, e.g. f(x) = x³ + x² − 12.5 changes sign between 2.025 and 2.035 (or iteration); A1 2.03 m.
A probe of mass m kg is launched vertically upwards from the surface of the Earth with speed 4000 m s⁻¹. When the probe is at distance x m from the centre of the Earth, the gravitational force on it is mgR²/x² N, where R = 6.4 × 10⁶ m is the radius of the Earth and g = 10 m s⁻². Air resistance is ignored. The speed of the probe is v m s⁻¹.
- (a) Show that, while the probe is moving upwards, v dv/dx = −gR²/x². [1]
- (b) Find the greatest height above the Earth's surface reached by the probe. Give your answer in kilometres, correct to 3 significant figures. [5]
Show answer
- (a) Shown
B1 mv dv/dx = −mgR²/x², divide by m.
- (b) 914 km
M1 integrate: v²/2 = gR²/x + c; M1 at x = R, v = 4000: c = 8 × 10⁶ − gR; M1 v = 0: gR²/X = gR − 8 × 10⁶; A1 X = 7.314 × 10⁶ m; A1 height X − R = 914 km.
A particle of mass 2 kg moves along the positive x-axis. It is released from rest at the point where x = 2. When its displacement from the origin O is x m, the only force acting on it is an attraction towards O of magnitude 32/x³ N. Its speed is v m s⁻¹.
- (a) Show that v² = 4(4 − x²)/x². [3]
- (b) Find the speed of the particle when x = 1. [1]
- (c) Find the time taken for the particle to move from x = 2 to x = 1. [5]
Show answer
- (a) Shown
M1 2v dv/dx = −32/x³; M1 v²/2 = 8/x² + c; A1 c = −2 from v = 0 at x = 2, so v² = 16/x² − 4 = 4(4 − x²)/x².
- (b) √12 ≈ 3.46 m s⁻¹
B1 v² = 12, v = 2√3 = 3.46 m s⁻¹.
- (c) √3/2 ≈ 0.866 s
M1 dx/dt = −2√(4 − x²)/x (moving towards O); M1 t = ∫₁² x/(2√(4 − x²)) dx; M1 = [−(1/2)√(4 − x²)]₁²; A1 (1/2)√3; A1 0.866 s.
