AS & A Level Further Mathematics (9231)
Momentum 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.
Q1Oblique collisions of spheres · Core · Paper 3 style · 7 marksThe spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 2 kg, moving with speed 5 m s⁻¹, strikes sphere B, of mass 3 kg, which is at rest. Before impact, A's velocity makes an angle of 60° with the line of centres. The coefficient of restitution is 0.5. Find the speed of B after the collision. [4]
- (b) Find the speed of A after the collision and the angle its new direction makes with the line of centres. [3]
Show answer
- (a) 1.50 m s⁻¹
B1 component of A along line 5cos 60° = 2.5; M1 momentum 2(2.5) = 2v_A + 3v_B; M1 restitution v_B − v_A = 0.5(2.5); A1 v_B = 1.50.
- (b) 4.34 m s⁻¹ at 86.7° to the line of centres
B1 v_A = 0.250, perpendicular component unchanged 5sin 60° = 4.33; M1 speed √(v_A² + 4.33²); A1 4.34 m s⁻¹, 86.7°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Two spheres A and B each have mass m. B is at rest and A moves with speed 6 m s⁻¹ at 30° to the line of centres when they collide. The collision is perfectly elastic (e = 1). Find the velocity components of A and of B after the collision, along and perpendicular to the line of centres. [4]
- (b) Show that A and B move off at right angles to each other, and that kinetic energy is conserved. [2]
Show answer
- (a) A: 0 along, 3 m s⁻¹ perpendicular; B: 3√3 m s⁻¹ along the line, 0 perpendicular
B1 components of A: 6cos 30° = 3√3 along, 3 perpendicular; M1 momentum m(3√3) = mv_A + mv_B; M1 v_B − v_A = 3√3; A1 v_A = 0, v_B = 3√3.
- (b) Shown
B1 A moves perpendicular to the line, B along it; B1 ½m(27 + 9) = ½m(9) + ½m(27).
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Spheres A and B have equal mass. B is at rest. A moves with speed u in a direction making an angle α with the line of centres, where tan α = ¾. The coefficient of restitution is ⅔. Find, in terms of u, the speeds of A and B after the collision. [6]
Show answer
- (a) A: (√85/15)u; B: (2/3)u
B1 components 4u/5 along, 3u/5 perpendicular; M1 momentum 4u/5 = v_A + v_B; M1 restitution v_B − v_A = ⅔(4u/5); A1 v_A = 2u/15; A1 v_B = 2u/3; A1 speed of A = √((2u/15)² + (3u/5)²) = √85 u/15.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 0.4 kg, moving at 3 m s⁻¹, strikes sphere B, of mass 0.6 kg, at rest. A's velocity makes 50° with the line of centres and e = 0.8. Find the speed of B after the collision and the direction in which B moves. [4]
- (b) State the component of A's velocity perpendicular to the line of centres after the collision, and explain why it is unchanged. [2]
Show answer
- (a) 1.39 m s⁻¹ along the line of centres
B1 3cos 50° = 1.93; M1 0.4(1.93) = 0.4v_A + 0.6v_B; M1 v_B − v_A = 0.8(1.93); A1 v_B = 1.39, along the line of centres (B receives an impulse only along that line).
- (b) 2.30 m s⁻¹; the spheres are smooth so the impulse acts only along the line of centres
B1 3sin 50° value; B1 reason.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Smooth spheres A (mass 3 kg) and B (mass 1 kg) collide when the line of centres is parallel to the unit vector i. Just before impact A has velocity (4i + 3j) m s⁻¹ and B has velocity (−2i + j) m s⁻¹. The coefficient of restitution is 0.5. Find the velocities of A and B just after impact, in vector form. [5]
- (b) Find the magnitude of the impulse on B. [2]
Show answer
- (a) A: (1.75i + 3j) m s⁻¹; B: (4.75i + j) m s⁻¹
B1 j-components unchanged; M1 3(4) + 1(−2) = 3v_A + v_B; M1 v_B − v_A = 0.5(4 − (−2)); A1 v_A = 1.75; A1 v_B = 4.75.
- (b) 6.75 N s
M1 1 × (4.75 − (−2)); A1 6.75 N s.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass m, moving with speed u, collides with sphere B, of mass 2m, which is at rest. Before impact A's direction makes an angle of 45° with the line of centres. After the impact A moves perpendicular to the line of centres. Find the coefficient of restitution e. [4]
- (b) Find the fraction of the original kinetic energy lost in the collision. [3]
Show answer
- (a) e = ½
M1 momentum along line mu/√2 = mv_A + 2mv_B; M1 restitution v_B − v_A = e u/√2; M1 v_A = 0 gives v_B = u/(2√2) and v_B = eu/√2; A1 e = ½.
- (b) ¼
M1 KE after = ½m(u/√2)² + ½(2m)(u/(2√2))² = 3mu²/8; M1 loss = ½mu² − 3mu²/8 = mu²/8; A1 fraction ¼.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 3m, and sphere B, of mass m, are moving towards each other. Immediately before impact, A has speed 4u at 60° to the line of centres and B has speed 2u at 60° to the line of centres, each velocity having a component directed towards the other sphere. The coefficient of restitution is ⅓. Find the components of the velocities of A and B along the line of centres after the impact. [5]
- (b) Find the speed of each sphere after the impact and the angle between the direction of motion of B and the line of centres. [4]
Show answer
- (a) A: u, B: 2u (both in the original direction of A's component)
B1 along-line components 2u and −u (taking A→B as positive); M1 3m(2u) + m(−u) = 3mv_A + mv_B; M1 v_B − v_A = ⅓(2u + u); A1 v_A = u; A1 v_B = 2u.
- (b) A: √13 u, B: √7 u; B at 40.9°
B1 perpendicular components 2√3u (A) and √3u (B) unchanged; M1 speeds by Pythagoras; A1 √13u and √7u; A1 tan⁻¹(√3/2) = 40.9°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) On a smooth horizontal floor in a school science lab in Peshawar, sphere A of mass 1 kg moving at 8 m s⁻¹ strikes an identical sphere B at rest. A's velocity makes 30° with the line of centres and e = 0.6. Find the angle through which A's direction of motion is turned by the collision. [6]
Show answer
- (a) 40.9°
B1 components 4√3 along, 4 perpendicular; M1 momentum and M1 restitution along line; A1 v_A = 0.2 × 4√3 = 1.39; M1 new angle with line tan⁻¹(4/1.39) = 70.9°; A1 deflection 70.9° − 30° = 40.9°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Two smooth spheres A and B, each of mass m and radius a, lie on a smooth horizontal table. B is at rest. A moves with speed u along a straight line whose perpendicular distance from the centre of B is a. Show that, at impact, A's velocity makes an angle of 30° with the line of centres. [2]
- (b) The coefficient of restitution is ½. Find, in terms of u, the speeds of A and B after the impact. [5]
- (c) Find the angle between A's direction of motion after the impact and the line of centres. [2]
Show answer
- (a) Shown
M1 distance between centres at impact is 2a; A1 sin α = a/(2a) = ½, α = 30° (AG).
- (b) A: (√19/8)u; B: (3√3/8)u
M1 momentum along line m(√3u/2) = mv_A + mv_B; M1 v_B − v_A = ½(√3u/2); A1 v_A = √3u/8; A1 v_B = 3√3u/8; A1 speed of A √(3/64 + 1/4)u = √19u/8.
- (c) 66.6°
M1 tan β = (u/2)/(√3u/8) = 4/√3; A1 66.6°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) In a snooker club in Lahore, the cue ball (mass 0.17 kg) moving at 2.4 m s⁻¹ strikes a red ball of equal mass at rest. Model the balls as smooth spheres of equal radius. At impact the cue ball's velocity makes 25° with the line of centres, and e = 0.9. Find the speed of the red ball after the impact. [4]
- (b) Find the magnitude of the impulse between the balls. [2]
- (c) Find the speed of the cue ball after the impact. [2]
Show answer
- (a) 2.07 m s⁻¹
B1 2.4cos 25° = 2.18; M1 momentum; M1 restitution; A1 v_B = 0.95 × 2.18 = 2.07.
- (b) 0.351 N s
M1 I = 0.17 × 2.07; A1 0.351 N s.
- (c) 1.02 m s⁻¹
M1 v_A = 0.05 × 2.18 = 0.109, perpendicular component 2.4sin 25°; A1 1.02.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 2m, moving with speed u at 30° to the line of centres, strikes sphere B, of mass m, which is at rest. After the impact A moves at 45° to the line of centres. Find the exact value of the coefficient of restitution. [6]
Show answer
- (a) e = 2 − √3 (≈ 0.268)
B1 components √3u/2 along, u/2 perpendicular; M1 2m(√3u/2) = 2mv_A + mv_B; M1 v_B − v_A = e√3u/2; A1 v_A = (2 − e)√3u/6; M1 tan 45° = (u/2)/v_A so v_A = u/2; A1 2 − e = √3, e = 2 − √3.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere B, of mass 4 kg, moving at 3 m s⁻¹, strikes sphere A, of mass 2 kg, which is at rest. B's velocity makes 40° with the line of centres and the coefficient of restitution is 0.25. Find the speed of A after the collision. [4]
- (b) Find the speed of B after the collision and the angle through which its direction of motion is deflected. [4]
Show answer
- (a) 1.92 m s⁻¹
B1 3cos 40° = 2.30; M1 4(2.30) = 4v_B + 2v_A; M1 v_A − v_B = 0.25(2.30); A1 v_A = 1.92.
- (b) 2.35 m s⁻¹; deflected through 15.2°
A1 v_B along = 1.34; M1 speed √(1.34² + 1.93²) = 2.35; M1 angle with line tan⁻¹(1.93/1.34) = 55.2°; A1 deflection 15.2°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Smooth sphere A, of mass 2 kg, moves with velocity 5i m s⁻¹ and strikes smooth sphere B, of mass 3 kg, at rest. At impact the line of centres is parallel to the vector 3i + 4j. The coefficient of restitution is ½. Find the components of A's velocity along and perpendicular to the line of centres before impact. [3]
- (b) Find the velocities of A and B after the impact, in the form ai + bj. [4]
Show answer
- (a) Along: 3 m s⁻¹; perpendicular part: (3.2i − 2.4j) m s⁻¹
M1 unit vector (0.6i + 0.8j); A1 5i·(0.6i + 0.8j) = 3; A1 5i − 3(0.6i + 0.8j) = 3.2i − 2.4j.
- (b) A: (3.38i − 2.16j) m s⁻¹; B: (1.08i + 1.44j) m s⁻¹
M1 2(3) = 2v_A + 3v_B; M1 v_B − v_A = 1.5; A1 v_A = 0.3, v_B = 1.8; A1 vectors 0.3(0.6i + 0.8j) + 3.2i − 2.4j and 1.8(0.6i + 0.8j).
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 0.5 kg, moving at 6 m s⁻¹ at 40° to the line of centres, collides with sphere B, of mass 1.5 kg, which is moving along the line of centres towards A at 2 m s⁻¹. The coefficient of restitution is 0.4. Find the velocity components of A and B along the line of centres after impact. [5]
- (b) Find the loss of kinetic energy in the collision. [3]
Show answer
- (a) A: 2.33 m s⁻¹ reversed (away from B); B: 0.309 m s⁻¹ in A's original direction
B1 6cos 40° = 4.60; M1 0.5(4.60) − 1.5(2) = 0.5v_A + 1.5v_B; M1 v_B − v_A = 0.4(4.60 + 2); A1 v_A = −2.33; A1 v_B = 0.309.
- (b) 6.85 J
M1 only along-line components change; M1 [½(0.5)(4.60)² + ½(1.5)(2²)] − [½(0.5)(−2.33)² + ½(1.5)(0.309)²]; A1 6.85 J.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass m, moving with speed u, collides with sphere B, of mass km, which is at rest. A's velocity makes an angle α with the line of centres, where sin α = 0.6. The coefficient of restitution is 0.75. After the collision A moves at right angles to its original direction. Show that the component of A's velocity along the line of centres after impact is −0.45u (that is, A's component is reversed). [3]
- (b) Find the value of k. [4]
Show answer
- (a) Shown
B1 before: (0.8u, 0.6u) along/perpendicular; M1 after: (v, 0.6u), scalar product with original = 0; A1 0.8uv + 0.36u² = 0, v = −0.45u (AG).
- (b) k = 25/3
M1 m(0.8u) = mv + kmw; M1 w − v = 0.75(0.8u) = 0.6u; M1 substitute v = −0.45u: w = 0.15u, 0.8 = −0.45 + 0.15k; A1 k = 25/3.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass m, moving with speed u at 60° to the line of centres, strikes sphere B, of mass 3m, which is at rest. The coefficient of restitution is e. Show that A's velocity component along the line of centres is reversed by the collision if and only if e > ⅓. [4]
- (b) Given that e = ⅔, find the angle between the directions of motion of A before and after the collision. [4]
Show answer
- (a) Shown
M1 momentum m(u/2) = mv_A + 3mv_B; M1 v_B − v_A = eu/2; A1 v_A = (1 − 3e)u/8; A1 v_A < 0 ⇔ e > ⅓ (AG).
- (b) 38.2°
B1 v_A = −u/8; M1 velocity after (−u/8, √3u/2), speed 7u/8; M1 cos φ = [(−u/8)(u/2) + 3u²/4]/(u × 7u/8) = 11/14; A1 φ = 38.2°.
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass 2m, has velocity of magnitude 5u, with components 3u along the line of centres (towards B) and 4u perpendicular to it. Sphere B, of mass 3m, moves with speed 2u perpendicular to the line of centres. They collide; the coefficient of restitution is ½. Find the speed of B after the collision and the angle its direction makes with the line of centres. [6]
- (b) Find, in terms of m and u, the loss in kinetic energy. [3]
Show answer
- (a) 2.69u at 48.0°
M1 2m(3u) = 2mv_A + 3mv_B; M1 v_B − v_A = ½(3u); A1 v_A = 0.3u; A1 v_B = 1.8u; M1 combine with 2u perpendicular: √(1.8² + 2²)u = 2.69u; A1 tan⁻¹(2/1.8) = 48.0°.
- (b) 4.05mu²
M1 only the along-line components change; M1 ½(2m)(3u)² − [½(2m)(0.3u)² + ½(3m)(1.8u)²]; A1 9 − 0.09 − 4.86 = 4.05, so 4.05mu².
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A moves with speed u at an angle α to the line of centres and strikes an identical sphere B at rest. The coefficient of restitution is e (e < 1). After the impact A moves at an angle β to the line of centres. Show that tan β = 2tan α/(1 − e). [4]
- (b) Given that α = 20° and β = 45°, find e correct to 3 significant figures. [2]
- (c) Explain why β > α for every value of e with 0 ≤ e < 1. [1]
Show answer
- (a) Shown
M1 momentum u cos α = v_A + v_B; M1 v_B − v_A = eu cos α; A1 v_A = ½(1 − e)u cos α; A1 tan β = u sin α / v_A (AG).
- (b) 0.272
M1 1 − e = 2tan 20°; A1 e = 0.272.
- (c) 2/(1 − e) > 1 so tan β > tan α
B1 2/(1 − e) ≥ 2 > 1 and tan is increasing on (0°, 90°).
The spheres are smooth and have equal radii. The line of centres is the line joining their centres at the instant of impact.
- (a) Sphere A, of mass m, moving with speed u at 30° to the line of centres, strikes sphere B, of mass M, which is at rest. The coefficient of restitution is e. After impact, A moves at right angles to its original direction. Show that M/m = 4/(3e − 1). [6]
- (b) Hence find the least possible value of M/m and state the condition on e for such a deflection to be possible. [2]
Show answer
- (a) Shown
B1 before (√3u/2, u/2); M1 after (v, u/2) with (√3u/2)v + u²/4 = 0, v = −u/(2√3); M1 momentum m(√3u/2) = mv + Mw; M1 w − v = e√3u/2; M1 eliminate w: v = (m − eM)√3u/(2(m + M)); A1 (m − eM)/(m + M) = −⅓ gives 4m = M(3e − 1) (AG).
- (b) Least M/m = 2 (when e = 1); need e > ⅓
B1 3e − 1 > 0 so e > ⅓; B1 4/(3e − 1) is least when e = 1, giving 2.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere moving on a smooth horizontal floor at 8 m s⁻¹ strikes a vertical wall. Its direction makes 30° with the wall. The coefficient of restitution is 0.5. Find the speed of the sphere after the impact and the angle its new direction makes with the wall. [5]
Show answer
- (a) 7.21 m s⁻¹ at 16.1° to the wall
B1 parallel component 8cos 30° = 4√3 unchanged; B1 normal component 8sin 30° = 4; M1 after: 0.5 × 4 = 2; A1 speed √(48 + 4) = √52 = 7.21; A1 tan⁻¹(2/(4√3)) = 16.1°.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere of mass 0.2 kg moves on a smooth horizontal surface with velocity (6i − 4j) m s⁻¹ and hits a straight smooth wall which lies along the x-axis. The coefficient of restitution is 0.75. Find the velocity of the sphere after the impact. [2]
- (b) Find the magnitude of the impulse exerted by the wall and the kinetic energy lost. [3]
Show answer
- (a) (6i + 3j) m s⁻¹
M1 i-component unchanged, j-component reversed and multiplied by e; A1 6i + 3j.
- (b) 1.4 N s; 0.7 J
M1 0.2(3 − (−4)) = 1.4 N s; M1 ½(0.2)(36 + 16) − ½(0.2)(36 + 9); A1 5.2 − 4.5 = 0.7 J.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere approaches a straight vertical cushion with velocity components 3 m s⁻¹ parallel to the cushion and 4 m s⁻¹ perpendicular to it. The coefficient of restitution is 0.25. Find the speed of the sphere after impact and the angle between its path and the cushion. [4]
- (b) Find the percentage of kinetic energy lost. [2]
Show answer
- (a) √10 ≈ 3.16 m s⁻¹ at 18.4° to the cushion
B1 perpendicular component 0.25 × 4 = 1; M1 speed √(3² + 1²); A1 √10 = 3.16; A1 tan⁻¹(1/3) = 18.4°.
- (b) 60%
M1 (25 − 10)/25; A1 60%.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere strikes a smooth vertical wall with its path at 45° to the wall and rebounds with its path at 30° to the wall. Find the exact coefficient of restitution. [3]
- (b) Find the fraction of the sphere's kinetic energy lost in the impact. [3]
Show answer
- (a) e = 1/√3 (= √3/3)
M1 parallel component unchanged: v cos 45°; M1 tan 30° = e v sin 45° / (v cos 45°) = e; A1 e = 1/√3.
- (b) ⅓
M1 KE ∝ speed²; before v², after v²cos²45°(1 + tan²30°) = v²(½)(4/3) = ⅔v²; A1 loss ⅓v² relative to v², A1 fraction ⅓.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) On a smooth horizontal table, two straight smooth cushions meet at right angles at a corner. A sphere moving at 5 m s⁻¹ strikes one cushion and then the other. The coefficient of restitution for each impact is 0.6. Show that after both impacts the sphere moves parallel to its original path but in the opposite direction. [4]
- (b) Find the final speed and the fraction of the original kinetic energy remaining. [2]
Show answer
- (a) Shown
B1 axes along the cushions: velocity (a, b) with a² + b² = 25; M1 first impact (a, −0.6b)… i.e. one component reversed and scaled by 0.6; M1 second impact reverses and scales the other; A1 final velocity (−0.6a, −0.6b) = −0.6 × original (AG).
- (b) 3 m s⁻¹; 0.36
B1 0.6 × 5 = 3; B1 0.6² = 0.36.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A small smooth sphere falls vertically and hits a smooth plane inclined at 30° to the horizontal, with speed 6 m s⁻¹ just before impact. The coefficient of restitution is 0.5. Find the components of the sphere's velocity parallel and perpendicular to the plane immediately after impact. [3]
- (b) Find the speed of the sphere immediately after impact and the angle its velocity makes with the plane. [3]
Show answer
- (a) Parallel: 3 m s⁻¹ down the line of greatest slope; perpendicular: 2.60 m s⁻¹ away from the plane
B1 parallel component 6sin 30° = 3 unchanged; B1 normal component 6cos 30° = 3√3; B1 after: 0.5 × 3√3 = 1.5√3 = 2.60.
- (b) 3.97 m s⁻¹ at 40.9° to the plane
M1 √(3² + (1.5√3)²) = √15.75; A1 3.97; A1 tan⁻¹(1.5√3/3) = 40.9°.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere moving on a smooth horizontal table at 4 m s⁻¹ hits a straight smooth cushion and rebounds with speed 3 m s⁻¹. The coefficient of restitution is 0.5. Find the angle between the sphere's original path and the cushion. [5]
Show answer
- (a) 49.8°
B1 components 4cos α and 4sin α; B1 after: 4cos α and 2sin α; M1 16cos²α + 4sin²α = 9; M1 16 − 12sin²α = 9, sin²α = 7/12; A1 α = 49.8°.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A sphere hits a smooth wall with its path making an angle α with the wall. The coefficient of restitution is e. Show that the fraction of kinetic energy lost is (1 − e²)sin²α. [3]
- (b) Given that e = 0.6 and that 36% of the kinetic energy is lost, find α. [3]
Show answer
- (a) Shown
M1 components v cos α (unchanged) and v sin α → ev sin α; M1 loss ∝ v²sin²α − e²v²sin²α; A1 divide by v² (AG).
- (b) 48.6°
M1 0.64 sin²α = 0.36; A1 sin α = 0.75; A1 α = 48.6°.
The sphere is small and smooth, and the barrier (wall, cushion or plane) is smooth and fixed.
- (a) A ball of mass 0.15 kg, modelled as a small smooth sphere, slides at 10 m s⁻¹ across smooth horizontal ice and strikes a straight vertical barrier, its path making an angle α with the barrier. The coefficient of restitution is 0.64. After the impact the ball moves at right angles to its original path. Show that tan²α = 1/e and find α. [5]
- (b) Find the magnitude of the impulse exerted by the barrier on the ball. [2]
Show answer
- (a) α = 51.3°
B1 before (10cos α, −10sin α); B1 after (10cos α, 10e sin α); M1 scalar product zero: 100cos²α − 100e sin²α = 0; A1 tan²α = 1/e (AG); A1 tan α = 1.25, α = 51.3°.
- (b) 1.92 N s
M1 0.15 × 10sin α × (1 + 0.64); A1 1.92 N s.
- (a) Two smooth spheres A (mass 3 kg) and B (mass 2 kg), of equal radius, move towards each other along the same straight line with speeds 4 m s⁻¹ and 1 m s⁻¹ respectively, and collide directly. The coefficient of restitution is 0.6. Find the velocities of A and B after the collision. [4]
- (b) Find the kinetic energy lost in the collision. [2]
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- (a) A: 0.800 m s⁻¹, B: 3.80 m s⁻¹, both in A's original direction
M1 3(4) + 2(−1) = 3v_A + 2v_B; M1 v_B − v_A = 0.6(5) = 3; A1 v_A = 0.8; A1 v_B = 3.8.
- (b) 9.60 J
M1 (24 + 1) − (0.96 + 14.44); A1 9.60 J.
- (a) Three smooth spheres A, B and C, of equal radius and masses m, 2m and 4m, lie at rest in a straight line on a smooth horizontal surface, in that order. A is projected towards B with speed u. The coefficient of restitution between each pair is e. Find, in terms of e and u, the velocities of A and B after their collision. [4]
- (b) B then collides with C. Find the value of e for which B is brought to rest by this collision, and show that A is then also at rest. [4]
- (c) For this value of e, find the total kinetic energy lost in terms of m and u. [2]
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- (a) v_A = (1 − 2e)u/3, v_B = (1 + e)u/3
M1 mu = mv_A + 2mv_B; M1 v_B − v_A = eu; A1 v_A; A1 v_B.
- (b) e = ½; v_A = 0
M1 B–C collision: 2m v_B = 2m w_B + 4m w_C, w_C − w_B = e v_B; M1 w_B = (2 − 4e)v_B/6; A1 w_B = 0 ⇔ e = ½; A1 v_A = (1 − 1)u/3 = 0.
- (c) 3mu²/8
M1 v_B = u/2, w_C = (1 + e)v_B/3 = u/4; only C moves: ½(4m)(u/4)² = mu²/8; A1 loss ½mu² − mu²/8 = 3mu²/8.
