Teaching
Newton’s cradle
Pull back one ball and let go: a single ball swings out on the other side. Pull back two, and two swing out. This desk toy shows two conservation laws at work.
Controls
1 means a perfectly elastic collision. Below 1, some kinetic energy is lost in every collision.
- Mass of each ball: 50 g
- String length: 20 cm
- g = 9.8 m/s²
At impact
- Speed just before impact
- 0.725 m/s
- Speeds just after impact
- Ball 50.725 m/s
- Total momentum
- Before0.0362 kg·m/s
- After0.0362 kg·m/s
- Total kinetic energy
- Before0.0131 J
- After0.0131 J
The laws
- Speed at the lowest point, from conservation of mechanical energy on the way down
v = √(2 g L (1 − cos θ0))
- Momentum
p = m v, conserved in every collision
m v1 + m v2 = m v1′ + m v2′
- Kinetic energy
KE = ½ m v2, conserved only in a perfectly elastic collision
- Coefficient of restitution
e = (v2′ − v1′) / (v1 − v2)
- Two identical balls, the second at rest
v1′ = (1 − e) v1 / 2
v2′ = (1 + e) v1 / 2
for e = 1 the speeds swap
v1′ = 0
v2′ = v1
- Kinetic energy lost in that collision
the fraction (1 − e2) / 2
Assumptions
The balls are identical. The strings don’t stretch and their mass is negligible, and air resistance is neglected. Each collision is instantaneous and involves only two neighbouring balls, so collisions pass from ball to ball. A real Newton’s cradle is subtler, because its balls are touching; this simple model captures the essentials.
Try this
Pull back two balls. Why do you never see a single ball fly out twice as fast?
Show the explanation
a single ball at 2v would conserve momentum (m · 2v = 2 m v), but its kinetic energy would be ½ m (2v)² = 2 m v² instead of 2 × ½ m v² = m v². Together, the two laws require two balls to swing out, at speed v.
Set the coefficient of restitution to 0.8 and let go of one ball. What happens to momentum and to kinetic energy?
Show the explanation
total momentum stays the same; kinetic energy drops with every collision, and eventually all the balls move a little.
Change the release angle from 20° to 40°. Does the speed at the lowest point double?
Show the explanation
almost. Since 1 − cos θ₀ = 2 sin² (θ₀/2), v = 2 sin (θ₀/2) √(g L). And sin 20° / sin 10° ≈ 1.97, so the speed is multiplied by 1.97. For small angles, it is almost proportional to the angle.