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Impulse and momentum as vectors

Edexcel IAL Mathematics · WME02 Mechanics 2

M1 treated impulse and momentum along a line. In M2 they are vectors in i and j, so an impulse can change the direction of motion as well as the speed.

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Impulse-momentum principle

Momentum is the vector mv. An impulse I changes it:

I = mv - mu.

Work component by component. The magnitude of the impulse is |I| (N s), and its direction is the direction of the change in velocity.

Conservation of momentum

When two particles collide or join, with no external impulse, total momentum is conserved as a vector:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.

Equate the i parts and the j parts separately.

With calculus

A question may give r as a function of t (spec 1.4): differentiate to get the velocity at the moment of the impulse, then use I = mv - mu. Pearson's examiners note that a common slip is substituting into r instead of differentiating first.

Worked example

A ball of mass 0.5 kg has velocity (4i + 2j) m/s. It receives an impulse of (3i - 4j) N s. Find its new speed.

  1. v = u + I/m = (4i + 2j) + (6i - 8j) = 10i - 6j.
  2. Speed = sqrt(100 + 36) = 11.7 m/s.

Answer: 11.7 m/s

Key idea: I = mv - mu, as vectors. Momentum is conserved component by component. Speed is the magnitude of the final velocity.

Check yourself

  1. A particle of mass 2 kg changes velocity from (3i + 4j) m/s to (-i + 7j) m/s. Find the magnitude of the impulse.
    Show answer

    10 N s. I = 2((-i + 7j) - (3i + 4j)) = -8i + 6j, |I| = 10 N s.

  2. A 0.2 kg ball moving with velocity 5i m/s is struck and moves off with velocity (-3i + 4j) m/s. Find the magnitude of the impulse, to 3 significant figures.
    Show answer

    1.79 N s. I = 0.2(-8i + 4j) = -1.6i + 0.8j, |I| = 1.79 N s.

  3. A 3 kg particle moving with velocity (2i - j) m/s receives an impulse of (6i + 9j) N s. Find its speed afterwards, to 3 significant figures.
    Show answer

    4.47 m/s. v = (2i - j) + (2i + 3j) = 4i + 2j, speed sqrt(20) = 4.47 m/s.

  4. A particle of mass 2 kg moving with velocity (3i + j) m/s collides and coalesces with a particle of mass 1 kg moving with velocity (-i + 4j) m/s. Find the speed of the combined particle, to 3 significant figures.
    Show answer

    2.60 m/s. 3v = (6i + 2j) + (-i + 4j) = 5i + 6j, so v = (5i + 6j)/3 and the speed is sqrt(61)/3 = 2.60 m/s.

  5. A 0.5 kg ball moving at 6i m/s receives an impulse of 3j N s. Find the angle through which its direction of motion turns.
    Show answer

    45 °. v = 6i + 6j, which is at 45° to i.

  6. A particle of mass 0.5 kg has position r = t2 i + 3t j m. At t = 2 it receives an impulse of (-i + 2j) N s. Find its speed immediately afterwards, to 3 significant figures.
    Show answer

    7.28 m/s. v = 2t i + 3j = 4i + 3j at t = 2. New v = (4i + 3j) + (-2i + 4j) = 2i + 7j, speed 7.28 m/s.

  7. In which direction does an impulse act on a particle?
    • Always the direction of its final velocity
    • Always opposite to its initial velocity
    • The direction of the change in its momentum
    • Perpendicular to its motion
    Show answer

    The direction of the change in its momentum. I = mv - mu, so I is parallel to the change in momentum.

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Last updated 8 November 2026. Spotted a mistake? Tell us.