A ball on a smooth floor
A floor is a body of enormous mass that never moves, and its surface is the common tangent. A ball strikes it with speed \(u\) at an angle \(\theta\) to the normal, the vertical line at the point of impact. Resolve the velocity along the floor and along the normal, as in the last lesson.
Along the floor, a smooth floor gives no impulse, so the component \(u\sin\theta\) is unchanged. Along the normal, the floor stays at rest. So the ball's normal component is reversed and multiplied by \(e\): it arrives at \(u\cos\theta\) and leaves at \(eu\cos\theta\). Combine the two components to find the rebound.
The chips of Fig 6.1 show a pattern. With \(e = 1\) the ball leaves at the angle it arrived, like light from a mirror. With a smaller \(e\) the normal part shrinks, so the rebound is flatter and further from the normal. With \(e = 0\) the ball slides along the floor at \(u\sin\theta\).
A ball strikes a smooth floor at \(10\,\text{m/s}\), at \(53°\) to the normal, with \(e = 0.5\). Take \(\sin 53° = 0.8\) and \(\cos 53° = 0.6\). Find its speed just after the bounce, in m/s.
A ball strikes a smooth floor at speed \(u\), at an angle \(\theta\) to the normal, and \(e = 0\). Just after the impact the ball
Equal masses leave at a right angle
A snooker player strikes the cue ball into a still red ball, off-centre. The two balls separate, and the angle between their paths is always close to a right angle. This is no accident of the table. It follows from momentum and energy alone, for equal masses, one at rest, in an elastic collision.
\[ m\vec u = m\vec v_1 + m\vec v_2 \]
Why. momentum is conserved as a vector; the target starts at rest.
\[ \tfrac12 mu^2 = \tfrac12 mv_1^2 + \tfrac12 mv_2^2 \]
Why. the collision is elastic, so kinetic energy is conserved.
\[ u^2 = v_1^2 + v_2^2 + 2\,\vec v_1 \cdot \vec v_2 \]
Why. cancel \(m\) in line 1, and take the dot product of each side with itself.
\[ \vec v_1 \cdot \vec v_2 = 0 \]
Read it. line 2 says \(u^2 = v_1^2 + v_2^2\), so the extra term in line 3 must be zero.
The dot product of two vectors that are not zero vanishes only when they are perpendicular. So the two final velocities are at right angles. In the language of geometry, \(\vec u\), \(\vec v_1\) and \(\vec v_2\) form a right-angled triangle with \(\vec u\) as its hypotenuse. If the struck ball leaves at an angle \(\phi\) to the original direction, the striker leaves at \(90° - \phi\) on the other side.
The figure shows which ball takes what. The struck puck B leaves along the line of centres at \(u\cos\phi\), where \(\phi\) is the angle between the line of centres and A's path. The striker A keeps its tangential part and leaves at \(u\sin\phi\). A nearly head-on hit gives B a small \(\phi\) and most of the speed. A glancing touch leaves most of the speed with A.
A ball collides elastically and off-centre with an identical ball at rest. The struck ball leaves at \(60°\) to the incoming ball's original direction. The incoming ball leaves at
A carrom coin sliding at \(4\,\text{m/s}\) strikes an identical coin at rest, off-centre and elastically. The struck coin leaves at \(30°\) to the first coin's original direction. Find the speed of the first coin after the collision, in m/s.
Discs of different sizes
The direction of the struck body is the line of centres at the instant of contact. For two discs, this line is fixed by geometry. A disc of radius \(R\) slides along a path. The path passes a perpendicular distance \(d\) from the centre of a disc of radius \(r\) at rest. At contact the centres are \(R + r\) apart.
In the right triangle formed by the two centres and the path, the side opposite the angle \(\phi\) is \(d\), and the hypotenuse is \(R + r\). So \(\sin\phi = d/(R + r)\). If the discs have equal mass and the collision is elastic, the striker leaves at \(90° - \phi\) on the other side.
A disc of radius \(3\,\text{cm}\) slides at \(5\,\text{m/s}\) on a smooth table towards a disc of radius \(2\,\text{cm}\) and equal mass, at rest. The path of the first centre passes \(3\,\text{cm}\) from the centre of the second. The collision is elastic. Find the directions and speeds of both discs afterwards.
(a) At contact the centres are \(3 + 2 = 5\,\text{cm}\) apart, so \(\sin\phi = 3/5\) and \(\phi \approx 37°\). The struck disc leaves along the line of centres, \(37°\) from the original path.
(b) Along the line of centres, equal masses with \(e = 1\) swap components: the struck disc gets \(5\cos 37° = 4\,\text{m/s}\).
(c) The striker keeps its tangential component, \(5\sin 37° = 3\,\text{m/s}\), and leaves at \(53°\) on the other side of the path. Check: \(3^2 + 4^2 = 5^2\), so no kinetic energy is lost.
A disc of radius \(4\,\text{cm}\) moving on a smooth table strikes a disc of radius \(1\,\text{cm}\) at rest. The path of the first centre passes \(4\,\text{cm}\) from the centre of the second. At what angle to the original path does the struck disc move, in degrees?
Two bodies leave a collision at exactly \(90°\) to each other. Which conditions does the right-angle rule need?
In a cloud chamber, fast charged particles leave trails of tiny droplets that can be photographed. In the 1920s such photographs showed alpha particles striking helium nuclei, which have the same mass. The two tracks after each collision formed a fork at a right angle. The right angle told physicists that the collisions were elastic and that the two masses were equal.
- Multiplying the whole speed by \(e\) for a bounce off a floor.Only the normal component is multiplied by \(e\); the component along a smooth floor is kept.
- Using the right-angle rule for unequal masses or for inelastic collisions.The rule needs equal masses, a target at rest and an elastic collision.
- Taking \(\sin\phi = d/R\) or \(d/r\) for discs of different sizes.At contact the centres are \(R + r\) apart, so \(\sin\phi = d/(R + r)\).