Perfectly elastic: speed of separation equals speed of approach and no kinetic energy is lost.

The particles do not separate; they move together with a common velocity (coalesce).

Is momentum conserved when e<1?

Yes. Momentum is conserved in every collision; only kinetic energy is lost.

Write Newton's law of restitution as an equation.

v2-v1=e(u1-u2), where particle 2 is ahead in the positive direction.

What is a direct impact?

A collision in which the velocities are along the line of centres, before and after.

How do you find the kinetic energy lost in a collision?

Total \frac12mv^2 before minus total \frac12mv^2 after.

A calculation gives e=1.2. What does this tell you?

There is an error, since e\le1; kinetic energy cannot increase.

How do you find the impulse on one particle?

Its change in momentum, m(v-u).

Why must a positive direction be chosen?

Velocities are vectors: a particle moving the opposite way has a negative velocity in the equations.

Direct impact and Newton's law of restitutionEdexcel International A Level Further Maths: Flashcards

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State Newton's law of restitution for a direct impact.

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State Newton's law of restitution for a direct impact.: Speed of separation $=e\times$ speed of approach.
What range of values can the coefficient of restitution take?: $0\le e\le1$ .
What does $e=1$ mean?: Perfectly elastic: speed of separation equals speed of approach and no kinetic energy is lost.
What does $e=0$ mean?: The particles do not separate; they move together with a common velocity (coalesce).
State the conservation of momentum equation for two particles in a direct impact.: $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ .
Is momentum conserved when $e<1$ ?: Yes. Momentum is conserved in every collision; only kinetic energy is lost.
Write Newton's law of restitution as an equation.: $v_2-v_1=e(u_1-u_2)$ , where particle 2 is ahead in the positive direction.
What is a direct impact?: A collision in which the velocities are along the line of centres, before and after.
How do you find the kinetic energy lost in a collision?: Total $\frac12mv^2$ before minus total $\frac12mv^2$ after.
A calculation gives $e=1.2$ . What does this tell you?: There is an error, since $e\le1$ ; kinetic energy cannot increase.
How do you find the impulse on one particle?: Its change in momentum, $m(v-u)$ .
Why must a positive direction be chosen?: Velocities are vectors: a particle moving the opposite way has a negative velocity in the equations.

Exam questions on Direct impact and Newton's law of restitution

Particle $A$ of mass $2$ kg moves with speed $5$ m s $^{-1}$ along a straight line on a smooth horizontal surface and collides directly with particle $B$ of mass $3$ kg, which is at rest. The coefficient of restitution between the particles is $0.5$ .
Find the magnitude of the impulse exerted by $A$ on $B$ in the collision.2 marks
Particles $P$ and $Q$ , of mass $3$ kg and $5$ kg, move towards each other along the same straight line on a smooth horizontal surface with speeds $4$ m s $^{-1}$ and $2$ m s $^{-1}$ respectively. They collide directly. After the collision the direction of motion of $P$ is reversed and its speed is $1$ m s $^{-1}$ .
Find the total kinetic energy lost in the collision.2 marks
Two smooth spheres $A$ and $B$ of equal radii and masses $m$ and $3m$ lie on a smooth horizontal table. Sphere $A$ is projected with speed $u$ directly towards $B$ , which is at rest. The coefficient of restitution between the spheres is $e$ .
Show that the speed of $B$ immediately after the collision is $\frac{(1+e)u}{4}$ .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).