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A-Level · Further Mechanics 1 (Momentum, Impulse and Collisions Depth)

FP.FM5 Further Mechanics: Momentum, Impulse and Collisions Depth

EDEXCEL 9FM0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key results: momentum, impulse and Newton's law of restitution

Original content written for Revision Library.

Unless a question states otherwise, take g = 9.8 m/s^2, model each object as a particle, and take all strings as light and inextensible, all pulleys and surfaces as smooth, and all spheres as smooth (so there is no friction, and no tangential impulse, at any point of contact). The impulse-momentum principle states that the impulse J exerted on a particle of mass m equals its change in momentum, J = m*(v - u); when a force F varies with time t, the impulse delivered over an interval is the integral of F with respect to t over that interval. Provided no external horizontal force acts on a system of colliding or jerked particles, total momentum is conserved: m1*u1 + m2*u2 = m1*v1 + m2*v2, with velocities signed along the line of centres (or along the string). Newton's experimental law of restitution states that the speed of separation is e times the speed of approach, v2 - v1 = -e*(u2 - u1), where 0 <= e <= 1 is the coefficient of restitution between the two surfaces; e = 1 gives a perfectly elastic collision (no kinetic energy lost) and e = 0 gives a perfectly inelastic collision (the particles coalesce, or move off together immediately after a string jerks taut). The loss in kinetic energy due to a direct collision is m1*m2*(1-e^2)*(u1-u2)^2 / (2*(m1+m2)). When a sphere strikes a fixed smooth surface obliquely, the component of its velocity parallel to the surface is unchanged by the impact, while the component perpendicular to the surface obeys Newton's law of restitution.

A particle P of mass 0.4 kg is moving with velocity (3i - 2j) m/s when it receives an impulse of (2i + 5j) Ns.

(a)Find the velocity of P immediately after the impulse, giving your answer in the form (ai + bj) m/s.(3)

(b)Find the speed of P immediately after the impulse, giving your answer to 3 significant figures.(2)

(Total for Question 1 is 5 marks)

Two particles, A of mass 3 kg and B of mass 2 kg, lie at rest on a smooth horizontal plane, connected by a light inextensible string which is initially slack. Particle B is then projected directly away from A with speed 6 m/s, and travels in a straight line until the string becomes taut.

Figure (to be drawn): Particles A and B on a smooth horizontal line, connected by a slack string; B moves away from A until the string straightens and both move on together at a common velocity.

(a)Find the common speed of A and B immediately after the string becomes taut.(3)

(b)Find the magnitude of the impulsive tension in the string as it jerks taut.(3)

(c)Show that kinetic energy is lost when the string jerks taut, and find this loss.(3)

(Total for Question 2 is 9 marks)

Two smooth spheres, P and Q, of masses m and km respectively (where k > 0 is a constant), move towards each other in the same straight line with speeds u and v respectively (u > 0, v > 0), and collide directly. The coefficient of restitution between P and Q is e, where 0 ≤ e ≤ 1.

(a)Taking the direction of P's initial motion as positive, show that the velocities of P and Q immediately after the collision are v_P = [(1-ek)u - k(1+e)v]/(1+k) and v_Q = [(1+e)u + (e-k)v]/(1+k).(6)

(b)Hence show that, in the special case k = 1 and e = 1 (equal masses, a perfectly elastic collision), the two spheres exactly exchange velocities.(3)

(c)Hence show that, in the special case e = 0, the two spheres move on with a common velocity equal to (u-kv)/(1+k), and verify that this agrees with the value obtained directly from conservation of momentum alone.(2)

(Total for Question 3 is 11 marks)

A smooth sphere strikes a fixed smooth plane with speed 10 m/s, travelling at an angle of 30 degrees to the plane, and rebounds. The coefficient of restitution between the sphere and the plane is e = 0.6.

Figure (to be drawn): Sphere approaching a fixed plane at 30 degrees to the plane (60 degrees to the normal), showing the velocity resolved into components perpendicular and parallel to the plane, and rebounding at a shallower angle after impact.

(a)Find the components of the sphere's velocity, resolved perpendicular and parallel to the plane, immediately before impact.(3)

(b)State the parallel component of velocity immediately after impact, giving a reason, and find the perpendicular component of velocity immediately after impact.(3)

(c)Find the speed of the sphere, and the angle its path makes with the plane, immediately after impact, giving each answer to 3 significant figures.(4)

(d)Given that the mass of the sphere is 0.3 kg, find the magnitude of the impulse exerted on the sphere by the plane during the impact.(3)

(Total for Question 4 is 13 marks)

Three smooth spheres A, B and C, of masses 2 kg, 3 kg and m kg respectively, lie at rest in a straight line on a smooth horizontal table, with B between A and C. Sphere A is projected towards B with speed 8 m/s. The coefficient of restitution between A and B is 0.5, and the coefficient of restitution between B and C is 0.75.

Figure (to be drawn): Three spheres A, B and C in a line on a smooth table, with B between A and C; A is projected towards the stationary B and C.

(a)Find the velocities of A and B immediately after their collision.(5)

(b)Sphere B then collides directly with sphere C, which is also initially at rest. Given that B's velocity immediately after this second collision is 1.44 m/s (still in A's original direction of motion), find the value of m.(5)

(c)Determine, with justification, whether any further collisions occur between the three spheres.(3)

(Total for Question 5 is 13 marks)

Two smooth spheres, of masses m1 and m2, collide directly with signed velocities u1 and u2 respectively, separating with velocities v1 and v2 and coefficient of restitution e. Prove that the loss in kinetic energy due to the collision is m1*m2*(1-e²)*(u1-u2)² / (2*(m1+m2)).

(Total for Question 6 is 7 marks)

A ball is released from rest at a height of 4.9 m above smooth horizontal ground and allowed to bounce repeatedly. The coefficient of restitution between the ball and the ground is e = 0.5. Air resistance is negligible, and g = 9.8 m/s² throughout this question.

Figure (to be drawn): Ball falling from height 4.9 m, bouncing repeatedly to progressively lower heights, with the bounce interval times shrinking geometrically.

(a)Show that the ball takes exactly 1 second to fall to the ground for the first time, and find its speed immediately before this first impact.(3)

(b)Show that the time taken for the ball to rise to its highest point after the nth bounce, and fall back to the ground again, is 2*eⁿ seconds.(3)

(c)Hence find, as an exact number of seconds, the total time from the ball's release until it comes to rest.(3)

(Total for Question 7 is 9 marks)

Two particles, A of mass 4 kg and B of mass 6 kg, are connected by a light inextensible string which passes over a small smooth pulley fixed at the edge of a smooth horizontal table. Particle A rests on the table, and particle B hangs freely, with the string initially slack. Particle B is released from rest and falls freely (A remaining at rest, since the slack string exerts no force on it) through a distance of 5 m, at which point the string becomes taut.

Figure (to be drawn): Particle A on a smooth horizontal table connected via a string over a smooth pulley at the table's edge to particle B, which hangs and falls freely before the string jerks taut.

(a)Find, as an exact surd, the speed of B immediately before the string becomes taut.(2)

(b)Using the impulse-momentum principle, find the common speed of A and B immediately after the string jerks taut, and the magnitude of the impulsive tension in the string.(4)

(c)Find the loss in kinetic energy due to the string jerking taut, giving your answer to 3 significant figures.(2)

(Total for Question 8 is 8 marks)

Three particles, of masses 2 kg, 3 kg and 5 kg, have velocities (4i - j) m/s, (-2i + 3j) m/s and (i + 2j) m/s respectively.

(a)Find the total momentum of the system, giving your answer in the form (ai + bj) Ns.(3)

(b)Hence find the magnitude of the total momentum, giving your answer to 3 significant figures.(2)

(Total for Question 9 is 5 marks)

In snooker, a cue ball of mass 0.17 kg travelling at 3.6 m/s strikes a stationary red ball of the same mass directly. The coefficient of restitution between the balls is 0.9.

Figure (to be drawn): Cue ball approaching a stationary red ball of equal mass on a snooker table, with velocity arrows shown before and after the collision.

(a)Find the velocities of both balls immediately after the collision.(4)

(b)Calculate the percentage of the cue ball's original kinetic energy that is lost due to the collision, giving your answer to 3 significant figures.(3)

(Total for Question 10 is 7 marks)

A particle A of mass 3 kg, moving with speed 5 m/s, collides directly with a stationary particle B of mass 2 kg, and the two particles coalesce on impact to form a single combined particle. This combined particle then travels on and strikes a fixed smooth wall at right angles, rebounding with coefficient of restitution 0.4 between the combined particle and the wall.

Figure (to be drawn): Particle A approaching stationary particle B and coalescing, with the combined particle then travelling on to strike and rebound from a fixed wall.

(a)Find the common speed of the combined particle immediately after the first collision (with B).(2)

(b)Find the speed of the combined particle immediately after it rebounds from the wall.(2)

(c)Find the total loss in kinetic energy, from immediately before the first collision to immediately after the rebound from the wall.(3)

(d)Find, to 3 significant figures, the percentage of the original kinetic energy that has been lost overall.(2)

(Total for Question 11 is 9 marks)

During an impact lasting 0.5 seconds, the force F newtons exerted on a particle of mass 2 kg varies with time t seconds (0 ≤ t ≤ 0.5) according to F = 480*t - 960*t^2.

Figure (to be drawn): Sketch of the force-time graph F = 480t - 960t² over the interval 0 ≤ t ≤ 0.5 s: a single hump rising from zero, peaking, and returning to zero.

(a)Show that the impulse exerted on the particle during the whole impact is 20 Ns.(3)

(b)Given that the particle is initially at rest, find its speed immediately after the impact.(2)

(c)Find the time at which the force is greatest during the impact, and this greatest force, justifying that it is a maximum.(3)

(Total for Question 12 is 8 marks)

Mark scheme · FP.FM5 Further Mechanics: Momentum, Impulse and Collisions Depth

Question 1

(a) M1 rearranges the impulse-momentum principle to v = u + I/m
(a) A1 correct unsimplified substitution, e.g. (3i-2j) + (2i+5j)/0.4
(a) A1 v = (8i + 10.5j) m/s, cao
(a) Answer: v = (8i + 10.5j) m/s
(b) M1 speed = √8² + 10.5², using their velocity from part (a) (ft)
(b) A1 13.2 m/s (awrt 3 s.f.), cao
(b) Answer: 13.2 m/s (3 s.f.)

Question 2

(a) M1 uses conservation of momentum for the system: 3(0) + 2(6) = (3+2)v
(a) A1 12 = 5v (oe)
(a) A1 v = 2.4 m/s, cao
(a) Answer: 2.4 m/s
(b) M1 applies the impulse-momentum principle to A alone: impulsive tension = 3(2.4 - 0), using their v from part (a) (ft)
(b) A1 correct unsimplified expression, e.g. 3 x 2.4
(b) A1 7.2 Ns, cao
(b) Answer: 7.2 Ns
(c) M1 KE before the jerk = 0.5(2)(6²), since A is at rest and contributes no kinetic energy
(c) A1 KE after the jerk = 0.5(5)(2.4²), using their v from part (a) (ft)
(c) A1 loss = 36 - 14.4 = 21.6 J, cao (a positive loss confirms energy is lost)
(c) Answer: 21.6 J lost

Question 3

(a) M1 momentum equation (dividing throughout by m): u - kv = v_P + k*v_Q, since Q's initial velocity is -v
(a) M1 restitution equation: v_Q - v_P = -e(-v - u), simplifying to v_Q - v_P = e(u+v)
(a) dM1 eliminates v_P (or v_Q) by substitution, e.g. v_P = v_Q - e(u+v), into the momentum equation
(a) A1 correctly derives v_Q(1+k) = (1+e)u + (e-k)v, i.e. v_Q = [(1+e)u+(e-k)v]/(1+k)
(a) A1 back-substitutes to derive v_P = [(1-ek)u - k(1+e)v]/(1+k)
(a) A1 presents both final results together, fully simplified, cso
(a) Answer: v_P = [(1-ek)u - k(1+e)v]/(1+k); v_Q = [(1+e)u + (e-k)v]/(1+k)
(b) B1 substitutes k=1, e=1 into the formula for v_P: v_P = [(1-1)u - 1(2)v]/2 = -v
(b) B1 substitutes k=1, e=1 into the formula for v_Q: v_Q = [(2)u + (0)v]/2 = u
(b) B1 concludes that P now moves with the velocity Q originally had (-v), and Q now moves with the velocity P originally had (u), i.e. the spheres exactly exchange velocities, cso
(b) Answer: v_P = -v and v_Q = u, so P and Q exchange velocities exactly
(c) B1 substitutes e=0 into both formulae to obtain v_P = v_Q = (u-kv)/(1+k), cso
(c) B1 verifies against momentum conservation directly: m*u + km*(-v) = (m+km)*V gives V = (u-kv)/(1+k), identical to the value above
(c) Answer: Common velocity = (u-kv)/(1+k), matching the momentum-conservation result

Question 4

(a) M1 resolves the velocity using the 30 degree angle to the plane: perpendicular component = 10 sin(30), parallel component = 10 cos(30)
(a) A1 perpendicular component = 5 m/s (towards the plane)
(a) A1 parallel component = 5*√3 m/s (awrt 8.66 m/s), cao
(a) Answer: Perpendicular: 5 m/s (towards the plane); parallel: 5*√3 m/s (awrt 8.66 m/s)
(b) B1 states the parallel component is unchanged, since the plane is smooth so no impulse acts parallel to it: 5*√3 m/s (ft)
(b) M1 perpendicular component after impact = e x (perpendicular component before) = 0.6 x 5, using their value from part (a) (ft)
(b) A1 3 m/s (away from the plane), cao
(b) Answer: Parallel: 5*√3 m/s (unchanged); perpendicular: 3 m/s (away from the plane)
(c) M1 speed = √3² + (5*√3)², using their components from part (b) (ft)
(c) A1 9.17 m/s (awrt 3 s.f.), cao
(c) M1 angle to the plane: tan(α) = 3/(5*√3)
(c) A1 19.1 degrees (awrt 3 s.f.), cao
(c) Answer: Speed = 9.17 m/s (3 s.f.); angle to the plane = 19.1 degrees (3 s.f.)
(d) M1 recognises the impulse acts only perpendicular to the plane (the parallel component of velocity is unchanged, so no impulse acts parallel to the plane)
(d) M1 impulse = 0.3 x (3 - (-5)), using their perpendicular components from parts (a) and (b) (ft)
(d) A1 2.4 Ns, cao
(d) Answer: 2.4 Ns

Question 5

(a) M1 momentum equation: 2(8) + 3(0) = 2*v_A + 3*v_B
(a) A1 simplified to 2*v_A + 3*v_B = 16
(a) M1 restitution equation: v_B - v_A = -0.5(0-8)
(a) A1 simplified to v_B - v_A = 4
(a) A1 solves to v_A = 0.8 m/s and v_B = 4.8 m/s, both correct, cao
(a) Answer: v_A = 0.8 m/s; v_B = 4.8 m/s
(b) M1 momentum equation for the B-C collision, using B's speed from part (a) (ft): 3(4.8) + m(0) = 3(1.44) + m*v_C
(b) M1 restitution equation: v_C - 1.44 = 0.75(4.8 - 0)
(b) A1 v_C = 5.04 m/s
(b) dM1 substitutes v_C into the momentum equation to solve for m: 14.4 = 4.32 + m(5.04)
(b) A1 m = 2 kg, cao
(b) Answer: m = 2 kg
(c) B1 states A's velocity remains 0.8 m/s (unaffected by later events) while B moves at 1.44 m/s; since 0.8 < 1.44, A cannot catch up with B, so the gap between A and B does not close
(c) B1 states B's velocity after its collision with C is 1.44 m/s while C moves off at 5.04 m/s; since 1.44 < 5.04, B cannot catch up with C, so the gap between B and C does not close
(c) B1 concludes that, since each sphere in the line A, B, C moves more slowly than the one ahead of it, no further collisions occur between any pair
(c) Answer: No further collisions occur, since 0.8 < 1.44 < 5.04 with the spheres in the order A, B, C

Question 6

M1 writes the loss as 0.5*m1(u1-v1)(u1+v1) + 0.5*m2(u2-v2)(u2+v2), using the difference-of-two-squares factorisation of u1²-v1² and u2²-v2²
M1 uses momentum conservation m1*u1+m2*u2=m1*v1+m2*v2 to show m1(u1-v1) = m2(v2-u2)
A1 substitutes to reduce the loss to 0.5*m1(u1-v1)*[(u1+v1)-(u2+v2)]
M1 uses the restitution equation v1-v2=-e(u1-u2) to simplify (u1+v1)-(u2+v2) = (u1-u2)+(v1-v2) = (1-e)(u1-u2)
dM1 solves the momentum and restitution equations simultaneously for v1 to show m1(u1-v1) = m1*m2*(1+e)*(u1-u2)/(m1+m2)
A1 combines both results to obtain loss = m1*m2*(1-e²)*(u1-u2)² / (2*(m1+m2)), cso
B1 notes that since 0 ≤ e ≤ 1, (1-e²) ≥ 0, so the loss is always non-negative, confirming kinetic energy cannot be created in a collision
Answer: Loss = m1*m2*(1-e²)*(u1-u2)² / (2*(m1+m2)), which is always greater than or equal to 0 since 0 ≤ e ≤ 1

Question 7

(a) M1 uses h = 0.5*g*t² with h=4.9, g=9.8, i.e. 4.9 = 4.9*t²
(a) A1 t = 1 s, cso
(a) A1 speed = g*t = 9.8*1 = 9.8 m/s, cao
(a) Answer: t = 1 s; speed = 9.8 m/s
(b) M1 recognises the speed immediately after the nth bounce is eⁿ x 9.8, since each bounce multiplies the impact speed by e, starting from the first impact speed of 9.8 m/s
(b) M1 uses the fact that, in free flight under gravity with no resistance, the time to rise to the peak equals the time to fall back, so the total flight time is 2 x (speed just after the bounce)/g
(b) A1 = 2*(eⁿ*9.8)/9.8 = 2*eⁿ seconds, cso
(b) Answer: Time between the nth and (n+1)th bounce = 2*eⁿ seconds
(c) M1 total time = (time of first fall) + sum from n=1 to infinity of 2*eⁿ, using their results from parts (a) and (b) (ft)
(c) M1 recognises this as an infinite geometric series with first term 2e and common ratio e, and uses sum to infinity = 2e/(1-e)
(c) A1 total time = 1 + 2(0.5)/(1-0.5) = 1+2 = 3 seconds, cao
(c) Answer: Total time = 3 seconds

Question 8

(a) M1 uses v² = 2gh with g=9.8, h=5, i.e. v² = 98
(a) A1 v = √98 = 7*√2 m/s (awrt 9.90 m/s), cao
(a) Answer: 7*√2 m/s (awrt 9.90 m/s)
(b) M1 momentum conserved along the string during the jerk (only B has momentum beforehand): 6*(7*√2) = (4+6)*V, using their speed from part (a) (ft)
(b) A1 V = 4.2*√2 m/s (awrt 5.94 m/s), cao
(b) M1 impulsive tension found by applying the impulse-momentum principle to A alone: impulse = 4*(4.2*√2 - 0), using their V (ft)
(b) A1 16.8*√2 Ns (awrt 23.8 Ns), cao
(b) Answer: V = 4.2*√2 m/s (awrt 5.94 m/s); impulsive tension = 16.8*√2 Ns (awrt 23.8 Ns)
(c) M1 KE before = 0.5(6)(7*√2)² [A contributes 0]; KE after = 0.5(10)(4.2*√2)², using their V from part (b) (ft)
(c) A1 loss = 294 - 176.4 = 118 J (awrt 3 s.f.; exact value 117.6 J), cao
(c) Answer: 118 J (3 s.f.); exact value 117.6 J

Question 9

(a) M1 attempts the component-wise sum of m*v for all three particles
(a) A1 i-component: 2(4)+3(-2)+5(1) = 7
(a) A1 j-component: 2(-1)+3(3)+5(2) = 17, giving total momentum (7i+17j) Ns
(a) Answer: (7i + 17j) Ns
(b) M1 magnitude = √7² + 17², using their values from part (a) (ft)
(b) A1 18.4 Ns (awrt 3 s.f.), cao
(b) Answer: 18.4 Ns (3 s.f.)

Question 10

(a) M1 momentum equation: 0.17(3.6) + 0.17(0) = 0.17*v1 + 0.17*v2, simplifying (dividing by 0.17) to v1+v2=3.6
(a) M1 restitution equation: v2-v1 = -0.9(0-3.6), simplifying to v2-v1=3.24
(a) A1 cue ball velocity v1 = 0.18 m/s, cao
(a) A1 red ball velocity v2 = 3.42 m/s, cao
(a) Answer: Cue ball: 0.18 m/s; red ball: 3.42 m/s
(b) M1 KE before = 0.5(0.17)(3.6²); KE after = 0.5(0.17)(0.18²+3.42²), using their velocities from part (a) (ft)
(b) A1 loss = 1.1016 - 0.996948 = 0.105 J (awrt 3 s.f.)
(b) A1 percentage lost = 9.50% (awrt 3 s.f.), cao
(b) Answer: 9.50% (3 s.f.)

Question 11

(a) M1 momentum equation: 3(5) + 2(0) = (3+2)v
(a) A1 v = 3 m/s, cao
(a) Answer: 3 m/s
(b) M1 speed after = e x (speed before) = 0.4 x 3, using their speed from part (a) (ft)
(b) A1 1.2 m/s (moving away from the wall), cao
(b) Answer: 1.2 m/s, away from the wall
(c) M1 KE initial (before any collision) = 0.5(3)(5²), since B is at rest
(c) M1 KE final (after the wall rebound) = 0.5(5)(1.2²), using their speeds from parts (a) and (b) (ft)
(c) A1 total loss = 37.5 - 3.6 = 33.9 J, cao
(c) Answer: 33.9 J
(d) M1 percentage = (33.9/37.5) x 100, using their total loss from part (c) (ft)
(d) A1 90.4% (awrt 3 s.f.), cao
(d) Answer: 90.4% (3 s.f.)

Question 12

(a) M1 sets up the impulse as the definite integral of F with respect to t from 0 to 0.5
(a) A1 correct antiderivative: 240*t² - 320*t³
(a) A1 evaluates to 240(0.25) - 320(0.125) = 60 - 40 = 20 Ns, cso
(a) Answer: Impulse = 20 Ns
(b) M1 impulse = m*v, i.e. 2v = 20, using their impulse from part (a) (ft)
(b) A1 v = 10 m/s, cao
(b) Answer: 10 m/s
(c) M1 differentiates and sets dF/dt=0: 480 - 1920*t = 0
(c) A1 t = 0.25 s (within the interval 0 ≤ t ≤ 0.5), and justifies it is a maximum since F is a downward-opening quadratic in t (the coefficient of t² is negative)
(c) A1 F_max = 480(0.25) - 960(0.25)² = 120 - 60 = 60 N, cao
(c) Answer: t = 0.25 s; greatest force = 60 N

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