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Momentum and impulse in one dimensionEdexcel A-Level Further Maths: Revision notes

Section 1

Momentum

The momentum of a particle of mass mm moving with velocity vv is mvmv. Momentum is a vector, so in one dimension its sign gives its direction: choose a positive direction at the start and keep to it. The unit is N s (equivalently kg m s−1^{-1}). A particle of mass 0.40.4 kg moving at −6-6 m s−1^{-1} has momentum −2.4-2.4 N s.

Key termsmomentum
Exam tip

Draw a diagram marking the positive direction before you write any equation, and put a sign on every velocity.

Section 2

Impulse and the impulse-momentum principle

The impulse of a constant force FF acting for time tt is I=FtI=Ft. The impulse-momentum principle states that the impulse equals the change in momentum: I=Ft=mv−mu.I=Ft=mv-mu. Impulse is also measured in N s. A ball of mass 0.40.4 kg arrives at 66 m s−1^{-1} and leaves the bat at 99 m s−1^{-1} in the opposite direction: I=0.4(9)−0.4(−6)=6I=0.4(9)-0.4(-6)=6 N s. If the contact lasts 0.0150.015 s, the average force is 60.015=400\frac{6}{0.015}=400 N.

Key termsimpulseimpulse-momentum principle
Common mistake

When a particle reverses direction, subtracting the speeds. mv−mumv-mu must use signed velocities, so the change is m(v+∣u∣)m(v+|u|) in magnitude.

Section 3

Conservation of momentum

When two particles collide, the impulses they exert on each other are equal and opposite (Newton's third law). The total momentum of the system therefore does not change, provided no external horizontal impulses act (a smooth surface, say): m1u1+m2u2=m1v1+m2v2.m_1u_1+m_2u_2=m_1v_1+m_2v_2. Take one direction as positive, write every velocity with its sign, and solve. A negative answer means that particle moves in the negative direction.

Key termsprinciple of conservation of momentum
Exam tip

After finding an unknown velocity, check it against the context: a particle cannot pass through the one it collided with.

Section 4

Direct collisions and coalescence

In a direct collision the particles move along the same line before and after, so every velocity is along that line and only the conservation equation (and impulses) are needed. In coalescence the particles join and move with one common velocity vv: m1u1+m2u2=(m1+m2)v.m_1u_1+m_2u_2=(m_1+m_2)v. Example: 22 kg at 66 m s−1^{-1} meets 33 kg at −2-2 m s−1^{-1} and they coalesce. 12−6=5v12-6=5v, so v=1.2v=1.2 m s−1^{-1} in the direction of the 22 kg particle.

Key termsdirect collisioncoalesce

Section 5

Finding impulses and forces from collisions

The impulse on one particle is its change in momentum, and the impulse on the other is equal and opposite. For AA (33 kg) moving at 66 m s−1^{-1} and rebounding at −1-1 m s−1^{-1} after hitting BB (55 kg, at rest): BB moves off at 4.24.2 m s−1^{-1} (from 18=−3+5v18=-3+5v), the impulse on BB is 5(4.2)=215(4.2)=21 N s, and on AA is 3(−1)−3(6)=−213(-1)-3(6)=-21 N s. If the collision lasts 0.030.03 s, the average force is 210.03=700\frac{21}{0.03}=700 N. Giving the direction is part of the answer.

Key termsaverage force
Common mistake

Giving a force or impulse as a bare magnitude when the question asks for direction. State the direction in words, or use the sign relative to your positive direction.

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Exam questions on Momentum and impulse in one dimension

  1. A cricket ball of mass 0.40.4 kg is moving horizontally at 66 m s−1^{-1} towards a batsman. It is hit straight back along the same line and leaves the bat with speed 99 m s−1^{-1}. The ball is in contact with the bat for 0.0150.015 s.
    The same impulse is given to a ball of mass 0.40.4 kg that is initially at rest. Find the speed it acquires.2 marks
  2. Particle AA, of mass 33 kg, moves at 66 m s−1^{-1} along a smooth horizontal surface and collides directly with particle BB, of mass 55 kg, which is at rest. After the collision AA moves with speed 11 m s−1^{-1} in the opposite direction to its original motion. The collision lasts 0.030.03 s.
    Find the magnitude and direction of the average force exerted by AA on BB during the collision.2 marks
  3. Particle PP, of mass 22 kg, moves at 66 m s−1^{-1} and particle QQ, of mass 33 kg, moves at 22 m s−1^{-1} in the opposite direction, along the same straight line on a smooth horizontal surface. They collide directly and coalesce to form a single particle RR.
    Find the speed and direction of RR immediately after the collision.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).