Questions on momentum and impulse

Question 1

Define impulse, state its units, and explain why a rebound needs a larger impulse than being stopped dead.

Accepted Answer

Impulse is force times the time it acts, measured in newton seconds, and equals the change in momentum. Stopping a body removes its momentum; reversing it removes that momentum and supplies the same again the other way, so the change is the sum of the two speeds rather than just the first. B1 for the definition, B1 for newton seconds, B1 for the rebound explanation.

Question 2

A ball of mass 0.2 kg travelling at 12 m/s is struck straight back at 18 m/s. The contact lasts 0.03 s. Find the impulse and the average force.

Accepted Answer

Taking the outgoing direction as positive, u = −12 and v = 18. Impulse = 0.2(18) − 0.2(−12) = 3.6 + 2.4 = 6 N s. Average force = 6/0.03 = 200 N. M1 for the change in momentum with signs, A1 for the impulse, A1 for the average force.

Question 3

A constant horizontal force of 50 N acts for 0.4 s on a particle of mass 2 kg that is at rest on a smooth horizontal table. Find the speed of the particle at the end of the 0.4 s.

Accepted Answer

Impulse = Ft = 50 × 0.4 = 20 N s, and it equals the change in momentum: 2v − 0 = 20, so v = 10 m/s. M1 for Ft = mv − mu, A1 for 10 m/s. The same answer comes from a = 25 m/s² and v = at, since the impulse-momentum principle is Newton's second law multiplied through by the time; the impulse route is the one that survives when the force is not constant.

Question 4

A particle of mass 0.5 kg rests on a smooth horizontal table. A horizontal force of 3t² newtons acts on it from t = 0 until t = 2 s. Find the impulse delivered by the force and the speed of the particle when t = 2.

Accepted Answer

The force varies, so the impulse is the area under the force-time graph: I = ∫3t² dt from 0 to 2 = [t³] from 0 to 2 = 8 − 0 = 8 N s. M1 for integrating the force between the two times, A1 for 8 N s. The particle started from rest, so 0.5v = 8 and v = 16 m/s. A1. The force reaches 12 N at t = 2, and 12 × 2 = 24 N s would be the impulse of a constant 12 N over the whole two seconds; the force was smaller than that for all but the last instant, and the average force was 8/2 = 4 N.

Question 5

A sphere of mass 5 kg moving at 3 m/s catches up with one of mass 4 kg moving at 1 m/s in the same direction. After the collision the 5 kg sphere moves at 1.8 m/s. Find the velocity of the other.

Accepted Answer

Take the original direction of motion as positive and draw a before and after diagram. Momentum before = 5(3) + 4(1) = 19 N s. After the collision 5(1.8) + 4v = 19, so 4v = 19 − 9 = 10 and v = 2.5 m/s in the same direction.
M1 for conservation of momentum, A1 for 19 N s before, M1 for the equation after the collision, A1 for 2.5 m/s.
The struck sphere has sped up and the striking one has slowed, and 2.5 > 1.8 means the spheres have separated.

Question 6

A stationary object of mass 8 kg explodes into two pieces of mass 3 kg and 5 kg. The 3 kg piece moves off at 6 m/s. Find the velocity of the other piece.

Accepted Answer

Total momentum before is zero, since nothing is moving. After: 3(6) + 5v = 0, so 5v = −18 and v = −3.6 m/s: the 5 kg piece moves at 3.6 m/s in the opposite direction. B1 for zero momentum before, M1 for the momentum equation after, A1 for the magnitude, A1 for the direction.

Question 7

Explain, in terms of impulse, why a longer contact time reduces the force in a collision, and give one everyday application.

Accepted Answer

The change in momentum is fixed by the speeds before and after, so the impulse is fixed. Since impulse is force times time, spreading the same impulse over a longer contact reduces the average force in proportion. B1 for the impulse being fixed, B1 for the inverse relation between force and time, B1 for an example. Crumple zones, airbags, thick landing mats and bending the knees on landing all work this way.

Question 8

A particle P of mass 3 kg lies at rest on a smooth horizontal table, joined to a particle Q of mass 2 kg by a light inextensible string which is initially slack. Q is then projected along the table directly away from P with speed 5 m/s. Find the common speed of the particles immediately after the string becomes taut, and the magnitude of the impulse transmitted through the string.

Accepted Answer

While the string is slack neither particle feels the other, so at the moment it tightens one is still at rest and the other is still travelling at 5 m/s. The jerk is a pair of equal and opposite impulses along the string, one on each particle, so they cancel for the pair and the total momentum is conserved through it. The string is inextensible, so afterwards the two particles must be moving with the same speed. Before: 2(5) + 3(0) = 10 N s. After: (2 + 3)v = 10, so v = 2 m/s.
For the impulse, apply the impulse-momentum principle to one particle on its own. Taking the particle that started at rest, the impulse it receives is 3(2) − 3(0) = 6 N s. The other particle checks it: 2(2) − 2(5) = −6 N s, the same size in the opposite direction, which is Newton's third law.
M1 for conservation of momentum through the jerk, A1 for 10 = 5v, A1 for v = 2 m/s, M1 for the change in momentum of one particle alone, A1 for 6 N s.
Momentum survives the jerk and kinetic energy does not: 25 J before against 10 J after, so 15 J goes in the sudden tightening.

Question 9

A particle of mass 3 kg moving at 4 m/s collides directly with a particle of mass 2 kg moving at 6 m/s towards it, and the two coalesce. Find the velocity of the combined particle, and the kinetic energy lost in the collision.

Accepted Answer

Take the 3 kg particle's direction as positive, so the velocities before are 4 and −6. Momentum before: 3(4) + 2(−6) = 12 − 12 = 0. M1 for the total momentum with signs. After coalescence the 5 kg particle has momentum 5v = 0, so v = 0: the combined particle is at rest. A1. Kinetic energy before was ½(3)(4²) + ½(2)(6²) = 24 + 36 = 60 J, and after it is zero, so 60 J is lost. M1 for both kinetic energies, A1 for 60 J. All 60 J goes into the deformation, sound and heat of the impact.

Question 10

A shell of mass 6 kg is moving horizontally at 10 m/s when it explodes into two pieces of masses 2 kg and 4 kg, both of which continue to move along the original line of motion. The 2 kg piece moves off at 25 m/s in the original direction. Find the velocity of the 4 kg piece, and the kinetic energy released by the explosion.

Accepted Answer

The explosion is internal to the shell, so momentum is conserved: 6(10) = 2(25) + 4v, giving 4v = 60 − 50 = 10 and v = 2.5 m/s in the original direction. M1 for conservation of momentum with the shell's momentum before, A1 for 2.5 m/s. Kinetic energy before: ½(6)(10²) = 300 J. After: ½(2)(25²) + ½(4)(2.5²) = 625 + 12.5 = 637.5 J. M1 for both totals, A1 for 637.5 J. The explosion has released 637.5 − 300 = 337.5 J. A1. Setting the momentum before to zero, as for a shell at rest, gives −12.5 m/s for the 4 kg piece.

Question 11

A ball of mass 0.3 kg falling vertically strikes horizontal ground at 8 m/s. The ground exerts an impulse of 3.6 N s on the ball. Find the speed with which the ball rebounds, and the height to which it rises, taking g = 9.8 m/s² and ignoring the weight of the ball during the impact.

Accepted Answer

Take upwards as positive, so the ball arrives at −8 m/s and leaves at v. Impulse = change in momentum: 3.6 = 0.3v − 0.3(−8) = 0.3v + 2.4, so 0.3v = 1.2 and v = 4 m/s. M1 for the impulse as mv − mu with the arrival velocity negative, A1 for 4 m/s. Rising to rest under gravity, v² = 2gh gives h = 16/(2 × 9.8) = 0.816 m. M1 for v² = 2gh, A1 for 0.816 m. The impulse has to remove the 2.4 N s the ball arrived with before it can supply any upward momentum, which is why only 1.2 N s of the 3.6 N s becomes rebound. Treating the arrival velocity as +8 gives a rebound of 20 m/s, faster than the ball arrived, which the ground cannot produce.

Question 12

A ball of mass 2 kg is dropped from rest at a height of 5 m and rebounds to a height of 1.8 m. It is in contact with the ground for 0.05 s. Find the impulse the ground exerts on it and the average force, first ignoring the weight during the impact and then allowing for it. Take g = 9.8 m/s².

Accepted Answer

Speed on landing = √(2 × 9.8 × 5) = 9.90 m/s downwards. Rebound speed = √(2 × 9.8 × 1.8) = 5.94 m/s upwards. Taking upwards as positive, the change in momentum is 2(5.94) − 2(−9.90) = 11.88 + 19.80 = 31.7 N s upwards. Ignoring the weight, the average force is 31.68/0.05 = 634 N. Allowing for it, the weight acts downwards throughout the contact and contributes an impulse of 2(9.8)(0.05) = 0.98 N s downwards. The ground must supply 31.68 + 0.98 = 32.66 N s, an average force of 32.66/0.05 = 653 N. M1 for a landing speed from the drop, A1 for 9.90 m/s, A1 for 5.94 m/s, M1 for the change in momentum with signs, A1 for 31.7 N s, A1 for 634 N, M1 for including the impulse of the weight, A1 for 653 N. The two answers differ by about 3%. Over so short a contact the weight is normally neglected, and either answer stands provided the assumption is stated. Momentum is not conserved for the ball alone here, because the ground is an external body.

Question 13

A ball of mass 0.5 kg moving on a smooth horizontal floor at 8 m/s strikes a fixed vertical wall. Immediately before the impact its direction of motion makes an angle of 60° with the wall; immediately after it the ball is moving at 6 m/s in a direction making an angle of 30° with the wall, its motion along the wall still being in the same sense as before. Find the magnitude of the impulse the wall exerts on the ball.

Accepted Answer

Impulse is a vector, so resolve everything into a component along the wall and a component perpendicular to it, and take the perpendicular direction away from the wall as positive.
Before the impact: 8 cos 60° = 4 m/s along the wall, and 8 sin 60° = 6.93 m/s into the wall, which is −6.93 m/s.
After the impact: 6 cos 30° = 5.20 m/s along the wall in the same sense, and 6 sin 30° = 3 m/s away from the wall.
The impulse is the change in momentum, taken one component at a time. Along the wall: 0.5(5.196 − 4) = 0.598 N s. Perpendicular to the wall: 0.5(3 − (−6.928)) = 0.5 × 9.928 = 4.964 N s. Combining them, the magnitude is √(0.598² + 4.964²) = 5.00 N s.
M1 for resolving the velocities parallel and perpendicular to the wall, A1 for the components before the impact, A1 for the components after it, M1 for the change in momentum component by component, with the perpendicular components opposite in sign, A1 for the two components of the impulse, M1 for combining them by Pythagoras, A1 for 5 N s.
The component along the wall has changed, so the impulse is not perpendicular to the wall: it acts at 83.1° to the wall, so the wall is rough. Against a smooth wall the component along the wall survives untouched, the impulse comes out perpendicular, and only the perpendicular components need handling at all. Losing the sign on the incoming perpendicular component is the standard error: it turns a change of 9.93 m/s into one of 3.93 m/s and drops the answer to 2.05 N s.

Question 14

A particle P of mass 2 kg moving at 5 m/s on a smooth horizontal surface collides directly with a particle Q of mass 3 kg which is at rest. After the collision Q moves with speed v m/s, and P continues to move in the same direction as before. Find the set of possible values of v.

Accepted Answer

Let w m/s be the speed of the striking particle after the collision, in its original direction. Conservation of momentum gives 2(5) = 2w + 3v, so 2w + 3v = 10 and w = (10 − 3v)/2. One equation in two unknowns is not a dead end here, because the physical situation supplies two inequalities.
The striking particle keeps moving forwards, so w > 0, which needs 10 − 3v > 0 and therefore v < 10/3.
It also cannot end up faster than the particle in front of it, or it would pass through it, so w ≤ v. That gives (10 − 3v)/2 ≤ v, hence 10 ≤ 5v and v ≥ 2.
Putting the two together, 2 ≤ v < 10/3.
M1 for conservation of momentum with both unknowns, A1 for 2w + 3v = 10, M1 for the condition that the striking particle does not overtake, A1 for v ≥ 2, A1 for v < 10/3.
At the lower endpoint both particles move off at 2 m/s, so they travel on together, which is a perfectly good collision and is why that end is included. At the upper one the striking particle would be brought to rest, and a particle at rest is not moving in its original direction, so the wording excludes it.

Question 15

A horizontal force acts on a particle of mass m, at rest on a smooth horizontal table, for a time T. At time t the force is kt(T − t) newtons, where k is a positive constant, so that it grows from zero and dies away to zero again at t = T. Show that the impulse delivered is kT³/6, and that if the impulse is I then k = 6I/T³. Show also that the greatest force exerted is 1.5 times the average force over the whole time T.

Accepted Answer

The impulse is the integral of the force over the time it acts: I = ∫kt(T − t) dt from 0 to T = k∫(Tt − t²) dt = k[Tt²/2 − t³/3] from 0 to T. M1 for integrating the force between the two times. That is k(T³/2 − T³/3) = kT³/6. A1. Setting this equal to I gives k = 6I/T³. A1.
The force kt(T − t) is a quadratic in t with roots 0 and T, so by symmetry its greatest value is at t = T/2, where it is k(T/2)(T/2) = kT²/4; differentiating gives k(T − 2t) = 0 at the same place. M1 for locating the maximum. Substituting k = 6I/T³: the peak is (6I/T³)(T²/4) = 3I/(2T). A1. The average force over the time T is the force that would deliver the same impulse in the same time, I/T, and 3I/(2T) divided by I/T is 1.5. A1.
The particle's final speed is I/m whatever the shape of the force, since only the area under the graph enters mv − mu. With k = 3 and T = 2 the force is 6t − 3t², the impulse is 4 N s, the peak 3 N and the average 2 N, which is the ratio 1.5 in numbers. Multiplying the peak force by the time overstates the impulse by half as much again, and multiplying the mean of the end values, zero, by the time gives nothing.

Question 16

The table records four balls, each struck once by a racket. Every velocity is measured along the same straight line, and one fixed direction along that line is taken as positive throughout. Use the table to determine which ball receives the greatest impulse from the racket, and show that the racket exerts its greatest average force on a different ball.

Accepted Answer

The impulse on a ball is its change in momentum, m(v − u), and the signs in the table have to be carried into that subtraction: for the first three balls u is negative, so subtracting it adds to v rather than taking away from it. M1 for using m(v − u) with the signs read from the table.
Ball A: 0.15(25 − (−20)) = 0.15 × 45 = 6.75 N s. Ball B: 0.25(18 − (−12)) = 0.25 × 30 = 7.5 N s. Ball C: 0.06(30 − (−30)) = 0.06 × 60 = 3.6 N s. Ball D: 0.40(19 − 5) = 0.40 × 14 = 5.6 N s. A1 for all four impulses. The largest of the four is ball B, at 7.5 N s. A1 for identifying B.
The average force is the impulse divided by the time over which it is delivered, and each ball has its own contact time. M1 for dividing each impulse by its own contact time. Ball A: 6.75/0.020 = 337.5 N. Ball B: 7.5/0.025 = 300 N. Ball C: 3.6/0.015 = 240 N. Ball D: 5.6/0.040 = 140 N. A1 for the four average forces. The largest is ball A, at 337.5 N, so the greatest average force is not exerted on the ball that receives the greatest impulse. A1 for the comparison and the conclusion.
Ball D is the one that never turns round, so its impulse comes from a difference of two positive velocities; treating its velocity before the strike as negative, as the first three demand, would give 0.40 × 24 = 9.6 N s and name the wrong ball.

ball	mass m / kg	velocity before u / m s⁻¹	velocity after v / m s⁻¹	contact time t / s
A	0.15	−20	25	0.020
B	0.25	−12	18	0.025
C	0.06	−30	30	0.015
D	0.40	5	19	0.040