Questions on impulse and momentum as vectors

Question 1

Explain why a particle whose speed is unchanged may still have received an impulse.

Accepted Answer

Momentum is a vector, so it changes if the direction changes even when the magnitude does not. The impulse equals the vector change mv − mu, which is non-zero whenever v and u point different ways. B1 for momentum being a vector, B1 for the impulse being the vector change.

Question 2

A particle of mass 3 kg moving with velocity (2i − 5j) m/s receives an impulse of (6i + 9j) N s. Find its new velocity and speed.

Accepted Answer

The change in velocity is the impulse divided by the mass: (2i + 3j) m/s. So the new velocity is (2i − 5j) + (2i + 3j) = (4i − 2j) m/s, with speed √(16 + 4) = √20 = 4.47 m/s. M1 for dividing the impulse by the mass, A1 for the new velocity, A1 for the speed.

Question 3

A particle of mass 0.5 kg has velocity (4i + 3j) m/s. Write down its momentum as a vector and find the magnitude of that momentum.

Accepted Answer

Momentum = mv = 0.5(4i + 3j) = (2i + 1.5j) N s, of magnitude √(4 + 2.25) = 2.5 N s. B1 for the vector, B1 for the magnitude. The magnitude is the mass times the speed, 0.5 × 5.

Question 4

A particle of mass 1.5 kg receives an impulse of (−3i + 6j) N s and immediately afterwards has velocity (i + 2j) m/s. Find its velocity immediately before the impulse.

Accepted Answer

I = mv − mu rearranges to mu = mv − I = 1.5(i + 2j) − (−3i + 6j) = (1.5i + 3j) + (3i − 6j) = (4.5i − 3j) N s. Dividing by 1.5, u = (3i − 2j) m/s. M1 for rearranging the principle as mu = mv − I, A1 for the momentum before, A1 for the velocity. Subtracting the impulse from the velocity without first multiplying by the mass, or adding it instead of subtracting, both give a wrong vector; the momentum triangle has mu and I nose to tail making mv, so mu is the final momentum with the impulse taken away.

Question 5

A particle of mass 0.4 kg has its velocity changed from (5i + 2j) m/s to (−i + 6j) m/s. Find the impulse and its magnitude.

Accepted Answer

Impulse = 0.4(−i + 6j) − 0.4(5i + 2j) = 0.4(−6i + 4j) = (−2.4i + 1.6j) N s. Its magnitude is √(5.76 + 2.56) = √8.32 = 2.88 N s. M1 for mv − mu, A1 for the impulse, A1 for its magnitude. The speed changed from √29 = 5.39 to √37 = 6.08, so quoting the mass times that difference would have given 0.28 N s: an order of magnitude wrong.

Question 6

A particle of mass 4 kg with velocity (2i + 3j) m/s coalesces with one of mass 6 kg with velocity (−3i + j) m/s. Find the common velocity and speed.

Accepted Answer

Momenta: 4(2i + 3j) = 8i + 12j and 6(−3i + j) = −18i + 6j. Total = (−10i + 18j) N s. Dividing by the combined 10 kg gives v = (−i + 1.8j) m/s, with speed √(1 + 3.24) = 2.06 m/s. M1 for the total momentum, A1 for (−10i + 18j), M1 for dividing by the combined mass, A1 for the velocity and speed.

Question 7

A particle is struck and its direction of motion changes but its speed does not. State what this tells you about the direction of the impulse.

Accepted Answer

Since |v| = |u|, the vectors mu and mv have the same length, so the triangle they form with the impulse is isosceles. The impulse is therefore perpendicular to the bisector of the angle between the old and new velocities, and in particular it is along neither of them unless the motion is exactly reversed. B1 for the isosceles momentum triangle, B1 for the perpendicular to the bisector, B1 for it lying along neither velocity.

Question 8

A particle of mass 2 kg moving with velocity (3i + 4j) m/s receives an impulse that leaves it moving in the direction of i with speed 6 m/s. Find the impulse.

Accepted Answer

The final velocity is 6i m/s, so the final momentum is 12i and the initial momentum is 2(3i + 4j) = 6i + 8j. The impulse is 12i − (6i + 8j) = (6i − 8j) N s, of magnitude √(36 + 64) = 10 N s. M1 for the final momentum, M1 for mv − mu, A1 for the impulse, A1 for its magnitude. The j component had to be removed entirely, so the impulse is not along the direction of travel.

Question 9

A particle of mass 0.5 kg is moving with velocity (6i + 8j) m/s when it receives an impulse of magnitude 5 N s in the direction of the vector (−3i + 4j). Find the velocity of the particle immediately afterwards and describe its motion.

Accepted Answer

The vector (−3i + 4j) has magnitude √(9 + 16) = 5, so the impulse is 5 × (−3i + 4j)/5 = (−3i + 4j) N s. M1 for turning the magnitude and direction into a vector, A1 for the impulse. Then v = u + I/m = (6i + 8j) + (−3i + 4j)/0.5 = (6i + 8j) + (−6i + 8j) = 16j m/s. M1 for u + I/m, A1 for 16j. The particle now moves parallel to j at 16 m/s, faster than its speed of 10 m/s before: the impulse had a component against its i motion, which it cancelled exactly, and a component along its j motion, which it increased. Treating the 5 N s as if it acted along the original velocity gives a speed of 20 m/s and no change of direction, which the given direction rules out.

Question 10

Particles P and Q move on a smooth horizontal plane. P has mass 4 kg and Q has mass m kg. Before they collide, P has velocity (5i + j) m/s and Q has velocity (−i + 2j) m/s; after the collision P has velocity (2i + 3j) m/s and Q has velocity (5i − 2j) m/s. Find the impulse exerted on P, and hence the value of m, verifying that both components give the same value.

Accepted Answer

Impulse on P = 4(2i + 3j) − 4(5i + j) = (8i + 12j) − (20i + 4j) = (−12i + 8j) N s. M1 for mv − mu, A1 for the impulse. By Newton's third law the impulse on Q is the negative of this, (12i − 8j) N s, and it equals m(v − u) for Q: m[(5i − 2j) − (−i + 2j)] = m(6i − 4j). M1 for applying the equal and opposite impulse to Q. The i components give 6m = 12, so m = 2; A1. The j components give −4m = −8, so m = 2 again; B1 for the check. Q's change of velocity has to be parallel to the impulse it received, and the data were consistent with that. Had the two components disagreed, the data would have described a collision in which no single mass conserved momentum in both components.

Question 11

Particles A and B, of masses 1 kg and 3 kg, lie on a smooth horizontal table joined by a slack light inextensible string. B is at rest, and at the instant the string becomes taut A is moving at 8 m/s in a direction making 60° with the string, away from B. Find the velocity of B, the speed of A immediately after the jerk, and the impulsive tension in the string.

Accepted Answer

Resolve A's velocity along and across the string: 8cos60° = 4 m/s along it and 8sin60° = 6.93 m/s across it. The impulsive tension acts along the string, so momentum along the string is conserved and, the string being inextensible, both particles leave with the same component v along it: 1(4) = (1 + 3)v, so v = 1 m/s. M1 for resolving and conserving along the string. B receives no impulse across the string, so B moves along the string at 1 m/s; A1. A keeps its 6.93 m/s across the string, so its speed is √(1² + (8sin60°)²) = √(1 + 48) = 7 m/s; A1. The impulsive tension is the impulse given to B: 3 × 1 = 3 N s; B1. A now moves at arctan(6.93/1) = 81.8° to the string, swung round from 60° towards the perpendicular, because the jerk took part of its motion along the string and none of its motion across it. Resolving into i and j instead of along and across the string gives two coupled equations with the impulse direction still unknown.

Question 12

Two particles P and Q, each of mass 2 kg, are moving on a smooth horizontal plane. Before they collide P has velocity (3i + 4j) m/s and Q has velocity (−i + 2j) m/s. After the collision P has velocity (i + 6j) m/s. Find the velocity of Q after the collision and the impulse exerted on each particle.

Accepted Answer

Momentum is conserved as a vector, so treat the i and j components separately.
Before: 2(3i + 4j) + 2(−i + 2j) = (6i + 8j) + (−2i + 4j) = (4i + 12j) N s.
After: P carries 2(i + 6j) = (2i + 12j), so Q carries (4i + 12j) − (2i + 12j) = (2i + 0j), giving Q the velocity i m/s.
Impulse on P = 2(i + 6j) − 2(3i + 4j) = (−4i + 4j) N s, of magnitude √32 = 5.66 N s.
Impulse on Q = 2(i) − 2(−i + 2j) = (4i − 4j) N s, the same magnitude in the opposite direction, as Newton's third law requires.
M1 for conserving momentum as a vector, A1 for the total before the collision, A1 for the velocity of Q, M1 for an impulse as mv − mu, A1 for the impulse on P, A1 for the impulse on Q.

Question 13

Particles P, of mass 1 kg, and Q, of mass 2 kg, move on a smooth horizontal plane with velocities (5i + j) m/s and (−i + 2j) m/s respectively, and collide. A student claims that after the collision P has velocity (−3i + 3j) m/s. Find the velocity that Q would then have, and by considering kinetic energy decide whether the claim can be correct.

Accepted Answer

Momentum is conserved as a vector. Before: 1(5i + j) + 2(−i + 2j) = (3i + 5j) N s. After, P would carry 1(−3i + 3j), so Q would carry (3i + 5j) − (−3i + 3j) = (6i + 2j), giving Q the velocity (3i + j) m/s. M1 for conserving momentum in components, A1 for Q's velocity.
Kinetic energy before: ½(1)(5² + 1²) + ½(2)(1² + 2²) = 13 + 5 = 18 J. M1 for ½m times the sum of the squared components, A1 for 18 J. After: ½(1)(3² + 3²) + ½(2)(3² + 1²) = 9 + 10 = 19 J. M1 for the kinetic energy after, A1 for 19 J and the conclusion.
A collision between two particles with no external impulse can conserve kinetic energy or lose some of it, but it cannot create any: 19 J > 18 J, so the claim cannot be correct. Conservation of momentum alone does not rule the claim out, because momentum can be conserved by any number of outcomes; the energy test is a separate condition, and it is the one that fails here. Comparing speeds rather than kinetic energies, or comparing P's energy on its own, does not settle the question, since one particle can gain energy from the other.

Question 14

Particles A, of mass m, and B, of mass km, lie on a smooth horizontal table joined by a slack light inextensible string. B is at rest, and at the instant the string becomes taut A is moving with speed u in a direction making an angle θ with the string, away from B. Show that the impulsive tension has magnitude kmu cos θ/(1 + k), and that the fraction of the kinetic energy lost in the jerk is k cos²θ/(1 + k).

Accepted Answer

Along the string A has component u cos θ and B has none. The tension acts along the string, so momentum along it is conserved and both particles leave with the same component v in that direction: mu cos θ = (m + km)v, so v = u cos θ/(1 + k). M1 for conserving momentum along the string with a common final component, A1 for v.
The impulsive tension is the impulse given to B, which starts from rest: I = kmv = kmu cos θ/(1 + k). A1.
Across the string there is no impulse, so A keeps its component u sin θ. The kinetic energy after the jerk is ½m(v² + u² sin²θ) + ½kmv² = ½mu² sin²θ + ½m(1 + k)v² = ½mu² sin²θ + ½mu² cos²θ/(1 + k). M1 for the kinetic energy after with A's across component kept and both particles sharing v.
The loss is ½mu² − that: ½mu²(1 − sin²θ) − ½mu² cos²θ/(1 + k) = ½mu² cos²θ (1 − 1/(1 + k)) = ½mu² cos²θ × k/(1 + k). A1 for the loss. Dividing by ½mu², the fraction lost is k cos²θ/(1 + k). A1.
With k = 1.5 and θ = 60° the fraction is 0.15, which is the 15 J of 100 J in the lesson's example. The across component contributes nothing to the loss: the jerk is a perfectly inelastic collision in the direction of the string only, and the fraction is the one-dimensional result k/(1 + k) scaled by cos²θ, the share of the kinetic energy that was along the string to begin with. Writing the energy after as ½(1 + k)mv² alone drops A's across component and predicts a loss that grows with θ instead of shrinking.

Question 15

A particle of mass 0.5 kg is moving with velocity (6i + 2j) m/s when it is struck by a bat. The impulse from the bat is parallel to the vector (−i + 2j), and immediately afterwards the particle is moving parallel to j. Find the impulse and its magnitude, and the speed of the particle immediately after the blow.

Accepted Answer

The direction of the impulse is known and its size is not, so write it as I = k(−i + 2j) for some k > 0. M1 for introducing the scalar multiple. Momentum before is 0.5(6i + 2j) = (3i + j), so momentum after is (3i + j) + k(−i + 2j) = (3 − k)i + (1 + 2k)j. Moving parallel to j means the i component of the momentum is zero: 3 − k = 0, so k = 3. M1 for setting the i component to zero, A1 for k = 3.
The impulse is (−3i + 6j) N s, of magnitude √(9 + 36) = 3√5 = 6.71 N s; A1. The momentum after is 7j, so the velocity is 7j/0.5 = 14j and the speed is 14 m/s; A1.
Assuming instead that the speed is unchanged, or that the impulse is perpendicular to the original velocity, adds a condition the question did not give and contradicts the direction that it did.