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Impulse and momentum as vectors questions
The same principle applied in two dimensions, where the i components and the j components are treated separately and each gives its own equation. Nothing new is required beyond keeping the components apart consistently.
7 original questions · 25 marks · the impulse and momentum as vectors notes · Further Mechanics 1
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Explain why a particle whose speed is unchanged may still have received an impulse.
Worked 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. Working with speeds rather than velocities loses that information entirely.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.
Worked 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.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.
Worked 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.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.
Worked 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.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.
Worked 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 never along either of them. B1 for the isosceles momentum triangle, B1 for the perpendicular to the bisector, B1 for it lying along neither velocity.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.
Worked 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.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.
Worked 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.
The two impulses summing to zero is the check worth making, and it says the same thing as conservation of momentum.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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