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A-Level Impulse and the conservation of momentum and energy
What the A-Level syllabus expects for Impulse and the conservation of momentum and energy, and how to practise it.
What the syllabus expects
- Take impulse as the area lying under a body's force–time graph and apply that in calculations
- Give the statement of the principle of momentum conservation
- Use momentum conservation on straightforward one-dimensional two-body encounters, whether inelastic or perfectly elastic
Scope: the coefficient of restitution is not required - Recognise that in a perfectly elastic two-body collision the closing speed before impact matches the separating speed afterward
- Understand that a closed system keeps its total momentum through any interaction, though its kinetic energy generally shifts
How it's examined
Questions on this topic most often ask you to find, describe, determine, explain. About 3% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (4 marks)
Consider the three-ball chain of perfectly elastic collisions in which ball A hits ball B, and ball B then drives ball C forward. Hence, expressing your answer using m_A and m_C, determine the value of m_B that maximises the fraction F.
Show the worked answer
For an elastic collision the struck ball's speed is v' = [2m_hit/(m_hit+m_struck)]·v. A→B: v_B = 2m_A/(m_A+m_B)·v_0. B→C: v_C = 2m_B/(m_B+m_C)·v_B = 4m_A m_B /[(m_A+m_B)(m_B+m_C)]·v_0. F (velocity, and hence energy, fraction to C) is maximised when v_C is maximal, i.e. when g(m_B) = m_B/[(m_A+m_B)(m_B+m_C)] is a maximum. Writing (m_A+m_B)(m_B+m_C)/m_B = m_A + m_C + m_B + m_A m_C/m_B, set derivative to zero: 1 - m_A m_C/m_B² = 0, giving m_B² = m_A m_C. So m_B = √(m_A m_C).
Example 2 (3 marks)
A ball A of mass m_A, moving with velocity u_A, strikes a stationary ball B of mass m_B in a head-on, perfectly elastic collision. After impact, balls A and B travel along the line joining their centres with velocities v_A and v_B respectively. Prove that, however small the mass of ball B is made, the speed it gains after impact can never be greater than 2u_A.
Show the worked answer
Conservation of momentum: m_A u_A = m_A v_A + m_B v_B. For a perfectly elastic head-on collision the relative velocity of approach equals the relative velocity of separation: u_A = v_B - v_A, i.e. v_A = v_B - u_A. Substituting: m_A u_A = m_A(v_B - u_A) + m_B v_B => 2 m_A u_A = (m_A + m_B) v_B, so v_B = 2 m_A u_A / (m_A + m_B) = 2 u_A / (1 + m_B/m_A). Since m_B > 0, the denominator (1 + m_B/m_A) > 1, so v_B < 2 u_A. As m_B is made ever smaller, m_B/m_A -> 0 and v_B -> 2 u_A but never exceeds it. Hence B's speed can never be greater than 2 u_A.
Example 3 (2 marks)
(b) Describe what is meant by the zero-momentum frame (the centre-of-mass frame) and explain how using it makes the treatment of elastic collisions simpler.
Show the worked answer
The zero-momentum (centre-of-mass) frame is the reference frame that moves with the centre of mass of the system, so that the total linear momentum of the colliding bodies measured in it is zero. This simplifies elastic collisions because in this frame the two bodies always carry equal and opposite momenta, both before and after the collision. Combining this with conservation of kinetic energy in an elastic collision, each body simply reverses the direction of its velocity while keeping the same speed. The problem is therefore symmetric and one-dimensional in outcome, and results can be transformed back to the laboratory frame by adding the centre-of-mass velocity.
More worked questions on this topic
- The figure shows two rigid spheres, P and Q, each of mass m, moving towards one another with sp (2 marks)
- After an elastic collision a 0.20 kg bob on a 1.5 m string leaves at 2.857 m/s and swings up. F (3 marks)
- A 0.50 kg trolley moving at 2.0 m/s strikes a stationary 0.20 kg pendulum bob hung on a light 1 (3 marks)
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