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

FP.FM1 Further Mechanics: Momentum, Impulse and Collisions

AQA 7367 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A small ball of mass 0.5 kg is thrown so that it travels perpendicular to a smooth vertical wall. It strikes the wall with speed 8 m/s and rebounds, moving directly back along its original path, with speed 5 m/s.

Figure (to be drawn): Ball approaching a vertical wall and rebounding along the same line; velocity arrows labelled 8 m/s (incoming) and 5 m/s (outgoing, reversed direction).

(Total for Question 1 is 3 marks)

During a hockey training drill, Priya strikes a hockey ball of mass 0.16 kg that is travelling towards her at 12 m/s. The impact reverses the ball's direction of travel and sends it away from her at 18 m/s. The stick is in contact with the ball for 0.025 seconds.

(a)Calculate the magnitude of the impulse exerted on the ball by the stick.(3)

(b)Calculate the magnitude of the average force exerted on the ball by the stick during the impact.(2)

(Total for Question 2 is 5 marks)

Two trolleys, P of mass 5 kg and Q of mass 2 kg, are moving towards each other on a smooth horizontal track and collide directly. Immediately before the collision, P has speed 3 m/s and Q has speed 4 m/s, and they are moving in opposite directions. Immediately after the collision, P has velocity 0.6 m/s and Q has velocity 2 m/s, both now in the direction of P's original motion.

Figure (to be drawn): Trolleys P and Q on a straight track approaching each other, with velocity arrows labelled 3 m/s and 4 m/s before impact, and 0.6 m/s and 2 m/s (both in the same direction) after impact.

(a)By calculating the total momentum immediately before and immediately after the collision, show that momentum is conserved in this collision.(3)

(b)Calculate the coefficient of restitution between P and Q for this collision.(3)

(c)State, with a reason, whether this collision could be perfectly elastic.(1)

(Total for Question 3 is 7 marks)

Two smooth spheres, A and B, of mass 2 kg and 3 kg respectively, are free to move along the same straight line on a smooth horizontal table. Sphere A is projected towards B with speed 6 m/s, and at the same instant B is projected towards A with speed 4 m/s, so that the spheres collide directly. The coefficient of restitution between A and B is 0.5.

Figure (to be drawn): Spheres A and B on a horizontal line approaching each other, with velocity arrows labelled 6 m/s and 4 m/s before impact.

(a)Taking the direction of A's initial motion as positive, write down an equation for conservation of momentum for this collision.(2)

(b)Write down Newton's law of restitution (the restitution equation) for this collision.(2)

(c)Hence find the velocities of A and B immediately after the collision, stating clearly the direction in which each sphere moves.(5)

(d)Show that the collision results in a loss of kinetic energy, and find this loss.(3)

(Total for Question 4 is 12 marks)

This question tests recall of the key definitions used in collision problems.

(a)State Newton's experimental law of restitution for the direct impact of two smooth spheres, defining any symbols used.(2)

(b)State the range of possible values of e, and describe the physical significance of the values e = 0 and e = 1.(2)

(Total for Question 5 is 4 marks)

A smooth sphere A of mass 4 kg is moving with speed 5 m/s directly towards a smooth vertical wall. The coefficient of restitution between A and the wall is 0.4.

Figure (to be drawn): Sphere A approaching a vertical wall, rebounding, and then colliding with sphere B which is moving towards the wall from further away.

(a)Find the speed of A immediately after it rebounds from the wall.(2)

(b)After rebounding, A moves away from the wall and collides directly with a second smooth sphere, B, of mass 6 kg, which is moving towards the wall (that is, towards A) with speed 2 m/s. The coefficient of restitution between A and B is 0.75. Taking the direction away from the wall as positive, find the velocities of A and B immediately after this second collision.(6)

(c)Determine whether A and B collide again, giving a full reason for your answer.(3)

(Total for Question 6 is 11 marks)

Two small spheres, C and D, of equal mass m, move towards each other along the same straight line and collide directly. Immediately before the collision, C has velocity 8 m/s and D has velocity -2 m/s. Immediately after the collision, C has velocity 1 m/s and D has velocity 5 m/s.

(a)Verify that this is consistent with conservation of momentum, and find the coefficient of restitution e for this collision.(4)

(b)Given that m = 0.8 kg, calculate the loss in kinetic energy due to the collision.(2)

(Total for Question 7 is 6 marks)

Two railway trucks, P of mass 2400 kg travelling at 3 m/s, and Q of mass 3600 kg initially at rest, collide and couple together so that they move with a common velocity immediately after the collision (that is, e = 0 for this impact).

Figure (to be drawn): Truck P moving towards stationary truck Q, then both trucks moving together at a common velocity after coupling.

(a)Find the common velocity of the trucks immediately after the collision.(3)

(b)Calculate the loss in kinetic energy due to the collision.(3)

(c)Let m1 be the mass of a moving particle that collides directly with, and coalesces with, a second particle of mass m2 which is initially at rest (e = 0). Show that the fraction of the initial kinetic energy lost in the collision is m2 / (m1 + m2), and verify this general result using your answers to parts (a) and (b).(3)

(Total for Question 8 is 9 marks)

A ball is dropped from rest from a height of 2.5 m onto smooth, horizontal ground. The coefficient of restitution between the ball and the ground is e = 0.6. Air resistance may be ignored, and g = 9.8 m/s² throughout this question.

Figure (to be drawn): Ball falling from height 2.5 m, then bouncing to progressively lower heights after each impact with the ground.

(a)Find the speed of the ball immediately before it first strikes the ground.(2)

(b)Find the height to which the ball rebounds after the first bounce.(3)

(c)Show that the height reached after the nth bounce is h_n = 2.5 e²ⁿ, and hence find the height reached after the third bounce, giving your answer to 3 significant figures.(4)

(Total for Question 9 is 9 marks)

Two smooth spheres, of masses m1 and m2, collide directly. During the collision each sphere exerts a force on the other. By using Newton's third law together with the impulse-momentum principle, prove that the total momentum of the two spheres is conserved during the collision, stating clearly any assumption you make.

(Total for Question 10 is 4 marks)

Two smooth spheres, E and F, of masses 3 kg and 5 kg respectively, move towards each other on a smooth horizontal plane and collide directly. Before the collision, E has speed 2sqrt(3) m/s and F has speed √3 m/s, moving in opposite directions. The coefficient of restitution between E and F is e = 1/3.

Figure (to be drawn): Spheres E and F on a horizontal line approaching each other, with velocity arrows labelled 2sqrt(3) m/s and √3 m/s.

(a)Taking E's initial direction of motion as positive, find the exact velocities of E and F immediately after the collision, giving your answers in the form k √3, where k is a rational number.(6)

(b)Find, in exact form, the loss in kinetic energy due to the collision.(2)

(Total for Question 11 is 8 marks)

Three smooth spheres, X, Y and Z, of masses 1 kg, 2 kg and 3 kg respectively, lie at rest in a straight line on a smooth horizontal table, with Y between X and Z. Sphere X is projected with speed 10 m/s towards Y, which is initially at rest. The coefficient of restitution between X and Y is 0.5.

Figure (to be drawn): Three spheres X, Y and Z in a line: X approaching stationary Y, then (after rebounding, if applicable) Y approaching stationary Z further along the line.

(a)Find the velocities of X and Y immediately after their collision.(5)

(b)Sphere Y then goes on to collide directly with sphere Z, which is also initially at rest. The coefficient of restitution between Y and Z is 0.8. Find the velocities of Y and Z immediately after this second collision.(5)

(c)Explain why sphere Y must go on to collide with sphere X a second time.(3)

(Total for Question 12 is 13 marks)

A squash ball of mass 24 grams is served horizontally at a smooth vertical wall with speed 30 m/s. It rebounds and returns along the same horizontal line. The coefficient of restitution between the ball and the wall is e = 0.75.

Figure (to be drawn): Squash ball approaching a vertical wall and rebounding along the same line, with velocity arrows labelled 30 m/s (incoming) and 22.5 m/s (outgoing, reversed direction).

(a)Find the speed of the ball immediately after it rebounds from the wall.(2)

(b)Calculate the impulse exerted by the wall on the ball, stating an appropriate direction.(3)

(c)Calculate the percentage of the ball's kinetic energy that is lost in the impact, giving your answer to 3 significant figures.(2)

(Total for Question 13 is 7 marks)

Mark scheme · FP.FM1 Further Mechanics: Momentum, Impulse and Collisions

Question 1

M1 uses impulse = m(v - u), taking the direction towards the wall as positive so that u = 8 and v = -5 (oe)
A1 correct unsimplified expression, e.g. 0.5 x (-5 - 8)
A1 magnitude of impulse = 6.5 Ns (or 6.5 kg m/s), cao
Answer: 6.5 Ns (directed away from the wall)

Question 2

(a) M1 uses impulse = m(v - u), taking the ball's initial direction of travel as positive so u = 12 and v = -18 (oe)
(a) A1 correct unsimplified expression, e.g. 0.16 x (-18 - 12)
(a) A1 magnitude of impulse = 4.8 Ns, cao
(a) Answer: 4.8 Ns
(b) M1 uses force = impulse / time, F = 4.8 / 0.025 (ft on their impulse from part (a))
(b) A1 F = 192 N, cao
(b) Answer: 192 N

Question 3

(a) M1 momentum before, taking P's original direction as positive: 5(3) + 2(-4)
(a) A1 momentum before = 7 kg m/s
(a) B1 momentum after = 5(0.6) + 2(2) = 7 kg m/s, equal to the value before, so momentum is conserved (cso)
(a) Answer: Momentum before = momentum after = 7 kg m/s, so momentum is conserved
(b) M1 uses Newton's law of restitution, e = -(v_Q - v_P) / (u_Q - u_P)
(b) A1 correct unsimplified substitution, e = -(2 - 0.6) / (-4 - 3)
(b) A1 e = 0.2, cao
(b) Answer: e = 0.2
(c) B1 no, since e = 0.2 is not equal to 1 (a perfectly elastic collision requires e = 1, oe stating kinetic energy is not conserved)
(c) Answer: No, since e = 0.2 does not equal 1

Question 4

(a) M1 2(6) + 3(-4) = 2v_A + 3v_B (oe, correct masses and signed velocities substituted)
(a) A1 simplified to 2v_A + 3v_B = 0
(a) Answer: 2v_A + 3v_B = 0
(b) M1 v_B - v_A = -e(u_B - u_A) = -0.5(-4 - 6) (oe)
(b) A1 simplified to v_B - v_A = 5
(b) Answer: v_B - v_A = 5
(c) M1 rearranges one equation, e.g. v_A = -1.5v_B from part (a)
(c) dM1 substitutes into the equation from part (b) to form one equation in one unknown, v_B - (-1.5v_B) = 5
(c) A1 v_B = 2 m/s, cao
(c) A1 v_A = -3 m/s, cao (ft from their v_B if the method is correct)
(c) B1 correctly interprets the signs: A rebounds, moving at 3 m/s in the direction opposite to its initial motion; B continues moving at 2 m/s in A's original direction
(c) Answer: v_A = -3 m/s (i.e. 3 m/s reversed); v_B = 2 m/s (continuing in A's original direction)
(d) M1 KE before = 0.5(2)(6²) + 0.5(3)(4²)
(d) M1 KE after = 0.5(2)(3²) + 0.5(3)(2²), using their velocities from part (c) (ft)
(d) A1 loss = 60 - 15 = 45.0 J, cao
(d) Answer: Kinetic energy loss = 45.0 J

Question 5

(a) B1 correct statement, e.g. v2 - v1 = -e(u2 - u1) (oe), or 'speed of separation = e x speed of approach'
(a) B1 e correctly defined as the coefficient of restitution: a constant for the pair of colliding surfaces, independent of the masses and speeds involved
(a) Answer: v2 - v1 = -e(u2 - u1), where u1, u2 are the velocities before impact, v1, v2 the velocities after impact, and e is the coefficient of restitution between the two surfaces
(b) B1 0 ≤ e ≤ 1
(b) B1 e = 0 describes a perfectly inelastic collision (the spheres coalesce and move with a common velocity, the maximum possible kinetic energy is lost); e = 1 describes a perfectly elastic collision (no kinetic energy is lost)
(b) Answer: 0 ≤ e ≤ 1; e = 0 is perfectly inelastic (spheres coalesce), e = 1 is perfectly elastic (no KE lost)

Question 6

(a) M1 speed after = e x speed before = 0.4 x 5
(a) A1 2 m/s (moving away from the wall), cao
(a) Answer: 2 m/s, away from the wall
(b) M1 momentum equation: 4(2) + 6(-2) = 4v_A + 6v_B
(b) A1 simplified to 2v_A + 3v_B = -2 (oe, e.g. 4v_A+6v_B=-4)
(b) M1 restitution equation: v_B - v_A = -0.75(-2 - 2)
(b) A1 simplified to v_B - v_A = 3
(b) dM1 solves the two equations simultaneously
(b) A1 v_A = -2.2 m/s and v_B = 0.8 m/s, both correct, cao
(b) Answer: v_A = -2.2 m/s (2.2 m/s back towards the wall); v_B = 0.8 m/s (away from the wall)
(c) B1 recognises A is now moving back towards the wall at 2.2 m/s while B moves away at a constant 0.8 m/s, so A must strike the wall again before it could catch B
(c) M1 finds A's speed after rebounding from the wall a second time, 0.4 x 2.2
(c) A1 0.88 m/s, which exceeds B's constant speed of 0.8 m/s, so A gains on B and a second A-B collision must occur
(c) Answer: Yes, A and B collide again: A rebounds from the wall a second time at 0.88 m/s, faster than B's steady 0.8 m/s, so A catches up with B

Question 7

(a) M1 momentum before = m(8) + m(-2) = 6m; momentum after = m(1) + m(5) = 6m, equal for any m, so conserved
(a) A1 correctly states both totals equal 6m (oe)
(a) M1 uses e = -(v_D - v_C) / (u_D - u_C) = -(5 - 1) / (-2 - 8)
(a) A1 e = 0.4, cao
(a) Answer: Momentum before = after = 6m kg m/s (conserved); e = 0.4
(b) M1 KE before = 0.5(0.8)(8²) + 0.5(0.8)(2²); KE after = 0.5(0.8)(1²) + 0.5(0.8)(5²)
(b) A1 loss = 27.2 - 10.4 = 16.8 J, cao
(b) Answer: 16.8 J

Question 8

(a) M1 momentum equation: 2400(3) + 3600(0) = (2400+3600)v
(a) A1 7200 = 6000v (oe)
(a) A1 v = 1.2 m/s, cao
(a) Answer: 1.2 m/s
(b) M1 KE before = 0.5(2400)(3²) [Q at rest contributes 0]
(b) A1 KE after = 0.5(6000)(1.2²), using their v from part (a) (ft)
(b) A1 loss = 10800 - 4320 = 6480 J, cao
(b) Answer: 6480 J
(c) M1 uses momentum conservation to write the common velocity as v = m1 u / (m1 + m2)
(c) M1 forms KE before = 0.5 m1 u² and KE after = 0.5(m1+m2)v², and simplifies the loss to 0.5 m1 u² x [m2/(m1+m2)]
(c) A1 divides by the initial KE to obtain fraction lost = m2/(m1+m2) (cso), and verifies numerically: m2/(m1+m2) = 3600/6000 = 0.6, matching 6480/10800 = 0.6 from parts (a) and (b)
(c) Answer: Fraction lost = m2/(m1+m2); numerically 3600/6000 = 0.6, matching 6480/10800 = 0.6

Question 9

(a) M1 uses v² = u² + 2gh with u = 0, h = 2.5, g = 9.8
(a) A1 v = 7 m/s, cao (exact)
(a) Answer: 7 m/s
(b) M1 speed just after first bounce = e x 7 = 0.6 x 7 = 4.2 m/s (ft from part (a))
(b) M1 uses 0 = (4.2)² - 2gh to find the rebound height
(b) A1 h = 0.9 m, cao
(b) Answer: 0.9 m
(c) M1 recognises that since rebound speed = e x (impact speed) and height is proportional to (speed)², each successive rebound height is e² times the previous one
(c) A1 correctly derives h_n = 2.5 e²ⁿ (cso), noting h₁ = 2.5 e² = 0.9 agrees with part (b)
(c) M1 substitutes n = 3: h₃ = 2.5 x (0.6)⁶
(c) A1 h₃ = 0.117 m (awrt 3 s.f.), cao
(c) Answer: h_n = 2.5 e²ⁿ; h₃ = awrt 0.117 m

Question 10

M1 lets the spheres exert forces F and -F on each other during the contact time t (Newton's third law)
M1 applies the impulse-momentum principle to each sphere separately: -Ft = m1(v1-u1) and Ft = m2(v2-u2)
A1 adds the two equations to eliminate F: 0 = m1(v1-u1) + m2(v2-u2), rearranging to m1 u1 + m2 u2 = m1 v1 + m2 v2 (cso)
B1 states the assumption clearly: no external horizontal force acts on the system of two spheres during the (brief) collision, e.g. surfaces are smooth and the collision is direct along the line of centres
Answer: m1 u1 + m2 u2 = m1 v1 + m2 v2 (momentum conserved), assuming no external force acts on the two-sphere system during the collision

Question 11

(a) M1 momentum equation: 3(2sqrt(3)) + 5(-√3) = 3v_E + 5v_F
(a) A1 simplified to 3v_E + 5v_F = √3
(a) M1 restitution equation: v_F - v_E = -(1/3)(-√3 - 2sqrt(3))
(a) A1 simplified to v_F - v_E = √3
(a) dM1 solves the two equations simultaneously (e.g. substitutes v_F = v_E + √3)
(a) A1 v_E = -(1/2)√3 m/s and v_F = (1/2)√3 m/s, both exact and correct (oe equivalent surd forms), cao
(a) Answer: v_E = -(1/2)√3 m/s (i.e. (√3/2) m/s reversed); v_F = (1/2)√3 m/s
(b) M1 KE before = 0.5(3)(2sqrt(3))² + 0.5(5)(√3)²; KE after = 0.5(3)(√3/2)² + 0.5(5)(√3/2)², using their exact velocities from (a) (ft)
(b) A1 loss = 25.5 - 3 = 22.5 J, cao (exact)
(b) Answer: 22.5 J

Question 12

(a) M1 momentum equation: 1(10) + 2(0) = 1v_X + 2v_Y
(a) A1 simplified to v_X + 2v_Y = 10
(a) M1 restitution equation: v_Y - v_X = -0.5(0 - 10)
(a) A1 simplified to v_Y - v_X = 5
(a) A1 solves to v_X = 0 and v_Y = 5 m/s, both correct, cao
(a) Answer: v_X = 0 (X is brought to rest); v_Y = 5 m/s (in X's original direction)
(b) M1 momentum equation using Y's velocity from part (a): 2(5) + 3(0) = 2v_Y' + 3v_Z (ft)
(b) A1 simplified to 2v_Y' + 3v_Z = 10
(b) M1 restitution equation: v_Z - v_Y' = -0.8(0 - 5)
(b) A1 simplified to v_Z - v_Y' = 4
(b) A1 solves to v_Y' = -0.4 m/s and v_Z = 3.6 m/s, both correct, cao
(b) Answer: v_Y' = -0.4 m/s (Y rebounds, moving back towards X); v_Z = 3.6 m/s
(c) B1 states v_X = 0 from part (a), so with no external horizontal force acting on it, X remains at rest at its post-collision position indefinitely (unless struck again)
(c) B1 states v_Y' = -0.4 m/s from part (b), i.e. Y is moving back in the direction of X after its collision with Z
(c) B1 concludes that since Y moves directly towards the stationary X along the same line, with nothing between them, a second X-Y collision is inevitable
(c) Answer: Yes: X remains at rest (v_X = 0) while Y rebounds off Z at 0.4 m/s back towards X, so Y must eventually collide with X again

Question 13

(a) M1 speed after = e x speed before = 0.75 x 30
(a) A1 22.5 m/s, cao
(a) Answer: 22.5 m/s
(b) M1 impulse = m(v-u), taking the ball's original direction as positive so u=30, v=-22.5 (ft from part (a))
(b) A1 0.024 x (-22.5-30) = -1.26, magnitude 1.26 Ns
(b) A1 states the direction correctly: away from the wall (opposite to the ball's original direction of travel)
(b) Answer: 1.26 Ns, directed away from the wall
(c) M1 KE before = 0.5(0.024)(30²); KE after = 0.5(0.024)(22.5²) (or uses the shortcut fraction lost = 1 - e²)
(c) A1 43.8% (awrt 3 s.f.), cao
(c) Answer: 43.8% (3 s.f.)

Further Mechanics: Momentum, Impulse and Collisions - Worksheets, Questions and Revision