On a carrom board, a player flicks the striker at a coin that is not straight ahead. The striker glances off to one side, and the coin shoots away in a new direction, often into a corner pocket. Good players aim the striker at a point on the coin's edge, not at the coin's centre.
When bodies meet off-centre
In a head-on collision every velocity lay on one line. Now let two balls meet off-centre. They leave in different directions, and no single line holds all the velocities. Such a collision is an oblique collision. Momentum is a vector, so it must now be handled in two dimensions, as in the chapter on Motion in Two Dimensions.
Look at the instant of contact. The line of impact is the line at right angles to the two surfaces at the point of contact. For two spheres it passes through both centres; it is the line of centres of Lesson 1. The common tangent is the line through the point of contact at right angles to the line of impact.
Where the contact force points
The two directions of Fig 5.1 are chosen for a physical reason. Suppose the surfaces are smooth. A smooth surface cannot push along itself, because a push along the surface would be friction. So the contact force on each ball acts along the line of impact only. The impulse on B points along the line of impact, away from A, and the impulse on A points the opposite way.
This settles the carrom question at once. A coin at rest receives only the impulse along the line of impact. So it leaves along the line of impact, the line from the striker's centre through its own centre at the instant of contact. The striker's path does not decide it. Players therefore aim at the point of the coin's edge that makes this line point at the pocket.
Two smooth balls collide obliquely. During the impact, the force that each ball exerts on the other acts
Momentum along every direction
The contact forces are still a third-law pair, so the total momentum of the two balls is conserved. Now it is conserved as a vector. A vector equation holds when each of its components holds. So momentum is conserved along any direction we choose, and two perpendicular directions give two equations.
Take axes \(x\) and \(y\). Body 1 has velocity components \((u_{1x}, u_{1y})\) before and \((v_{1x}, v_{1y})\) after, and body 2 likewise.
\[ m_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}, \qquad m_1u_{1y} + m_2u_{2y} = m_1v_{1y} + m_2v_{2y} \]
To use these, resolve every velocity. A velocity \(u\) at an angle \(\alpha\) to the line of impact has the component \(u\cos\alpha\) along it and \(u\sin\alpha\) along the common tangent. The component along a direction always uses the cosine of the angle from that direction.
A ball moves at \(10\,\text{m/s}\) at \(53°\) to the line of impact. Take \(\sin 53° = 0.8\) and \(\cos 53° = 0.6\). Its components along the line of impact and along the common tangent are
Ball B moves at speed \(v\) along the \(x\) axis and strikes ball A, at rest, in an elastic collision. After the collision B moves at \(v/2\) along the \(y\) axis. Find the direction in which A moves, its speed, and the ratio of the masses.
(a) Let A leave at speed \(v_A\), at the angle \(\theta\) below the \(x\) axis. Momentum along \(x\): \(m_Bv = m_Av_A\cos\theta\). Along \(y\): \(0 = m_B\,\dfrac{v}{2} - m_Av_A\sin\theta\).
(b) Divide the second by the first: \(\tan\theta = \tfrac12\), so \(\theta \approx 26.6°\) below the \(x\) axis. Squaring and adding gives \(m_Av_A = \dfrac{\sqrt5}{2}\,m_Bv\).
(c) Kinetic energy: \(m_Bv^2 = m_B\dfrac{v^2}{4} + m_Av_A^2\), so \(m_Av_A^2 = \tfrac34 m_Bv^2\). Divide by \(m_Av_A\): \(v_A = \dfrac{3v}{2\sqrt5} \approx 0.67\,v\).
(d) The masses follow: \(m_A/m_B = \dfrac{\sqrt5}{2} \div \dfrac{3}{2\sqrt5} = \dfrac53\).
The smooth-ball rule
Along the common tangent, a smooth contact exerts no force at all. So neither ball receives any impulse along the tangent. Each ball therefore keeps its own tangential component of velocity through the collision. This is stronger than momentum conservation along the tangent, which is only about the total.
For smooth bodies, resolve every velocity along the line of impact and the common tangent. Along the tangent, each body's component is unchanged. Along the line of impact, treat the components as a head-on collision. Momentum is conserved, and the velocity of separation is \(e\) times the velocity of approach.
So the coefficient of restitution acts along the line of impact only. It compares the normal components of separation and of approach. The tangential components take no part in it. Applying \(e\) to the whole speed of a ball is the classic mistake of this topic.
A smooth ball A of \(1\,\text{kg}\) moves at \(10\,\text{m/s}\) and strikes an identical ball B at rest. At the instant of contact the line of impact makes \(37°\) with A's velocity, and \(e = 0.5\). Take \(\cos 37° = 0.8\) and \(\sin 37° = 0.6\). Find the velocities of both balls after the collision.
Resolve A's velocity. Along the line of impact \(n\): \(10\cos 37° = 8\,\text{m/s}\). Along the tangent \(t\): \(10\sin 37° = 6\,\text{m/s}\). B has no velocity at all.
Along \(t\): A keeps \(6\,\text{m/s}\) and B keeps \(0\). Along \(n\): \(8 = v_{An} + v_{Bn}\) and \(v_{Bn} - v_{An} = 0.5 \times 8 = 4\).
\(v_{Bn} = 6\,\text{m/s}\) and \(v_{An} = 2\,\text{m/s}\). So A has components \(2\,\text{m/s}\) along \(n\) and \(6\,\text{m/s}\) along \(t\), a speed of \(\sqrt{40} \approx 6.3\,\text{m/s}\).
B leaves along the line of impact at \(6\,\text{m/s}\). A leaves at \(6.3\,\text{m/s}\), about \(72°\) from the line of impact. The kinetic energy falls from \(50\,\text{J}\) to \(38\,\text{J}\).
Why the rule is needed
Count the unknowns in an oblique collision of two balls. Each ball has a final speed and a final direction, so there are four unknowns. Momentum gives two equations. Even for an elastic collision, kinetic energy adds only a third. The smooth-ball rule, with \(e\), supplies what conservation alone cannot.
A smooth ball moving at \(5\,\text{m/s}\) strikes an identical smooth ball at rest, elastically. At the contact, the line of impact makes \(60°\) with the moving ball's velocity. Find the speed of the struck ball after the collision, in m/s.
Two smooth balls collide obliquely with \(e = 0.6\). Which statements are true?
A smooth ball strikes another smooth ball obliquely. Explain why the first ball's velocity component along the common tangent is the same after the collision as before.
A change in a velocity component needs an impulse along that direction. The surfaces are smooth, so the contact force has no part along the common tangent; it acts along the line of impact only. With no tangential impulse, the tangential momentum of each ball, and so its tangential velocity, cannot change.
A carrom player who wants to pocket a coin draws an imaginary line from the pocket through the coin. The striker must touch the coin where this line leaves the coin, on the side away from the pocket. At contact the line of centres then points straight at the pocket, and the coin, pushed only along that line, runs into it.
- Applying \(e\) to the full speed of a ball in an oblique collision.\(e\) acts on the components along the line of impact only. The tangential components are unchanged for smooth bodies.
- Resolving with the sine and cosine swapped.The component along a line is \(u\cos\alpha\), with \(\alpha\) measured from that line.
- Expecting a struck ball at rest to move along the path of the striker.It moves along the line of impact, the only direction in which it receives an impulse.