Virginia Tech®home

Newton's Cradle

Newton's Cradle is a classic physics demonstration frequently seen as a desk decoration. This demonstration uses four or more suspended balls to demonstrate conservation of energy and conservation of momentum in a fairly elastic collision. While this demo is simple, it is easy to damage it by tangling the balls, so it must be packaged carefully.

Equipment

  • Newton's Cradle
  • Travel Box
    • The cradle must be returned to this box before being moved to prevent damage.

Back to Top

Physics Behind The Demo

This demo is used to show how conservation of momentum and energy works by using a series of swinging spheres. This is usually a system of five balls attached to a structure by two strings on either side. A ball on one end is lifted, and when it is released, strikes the other four balls. This collision creates a force through the other four balls and causes the ball on the other end to be pushed upward.[1][2]

Many ask why the cradle doesn't send two balls at half the speed when one ball is dropped. This does not happen because it would not conserve energy ($ \LARGE\mathbf{V} $ versus $\LARGE\mathbf{V^2} $!).

Elastic vs Inelastic Collisions

When considering collisions in physics, there are two types that can occur[:] elastic, and inelastic. In an elastic collision, both the momentum of the system and the kinetic energy of the system are conserved. That is:

$$ m_{in}v_{in} = m_{out}v_{out} \\ {and} \\ {{1} \over {2}} m_{in} {v_{in}}^{2} = {{1} \over {2}} m_{out} {v_{out}}^{2}$$

In an inelastic collision, only momentum is conserved. The kinetic energy put into the system does work on something, or is converted into another form of energy that can frequently escape the system. This does not violate the law of conservation of energy as no energy is actually missing, it has just been shifted elsewhere. Usually this energy ends up being used to deform an object permanently, such as a car crumpling in a crash.

Newton’s Cradle visualizes an elastic collision by allowing students to see the mass as a number of balls, and the speed as the height the balls travel.

Side Note: Gravitational Potential Energy and Energy Conservation

When the balls are lifted, they are given gravitational potential energy ($ \LARGE{mgh} $) that is completely converted to kinetic energy ($ \LARGE{{1} \over {2}} mv^{2} $) at the bottom of their swing. Setting these equations equal to one another, and using some algebra, we can see that the velocity is completely dependent on the height the balls are lifted.

$$ mgh_{top}+{{1} \over {2}} m{v_{top}}^{2}= mgh_{bottom}+{{1} \over {2}} m{{v_{bottom}}^{2}}$$

Since the velocity at the top of the swing (while it is being held) is 0 and the height at the bottom of the swing is also zero, this simplifies to:

$$ mgh_{top}+{{1} \over {2}} m \cdot 0 = mg \cdot 0+{{1} \over {2}} m{v_{bottom}}^{2} \rightarrow mgh_{top}={{1} \over {2}} m{v_{bottom}}^{2} $$

Solving for velocity $ \LARGE{v} $, we find:

$$ v_{out} = \sqrt{2gh}$$

Why the same number of balls?

Let’s examine the case of dropping two balls to collide with the remaining balls. As discussed above, when the balls collide, momentum must be conserved. So we know that:

$$ m_{2 balls} v_{i}=m_{? balls} v_{f} $$

Trying to solve anything we get something like:

$$ v_{f}= {m_{2 balls} \over m_{? balls}} v_{i} $$

This leaves us with two unknowns, the final mass and the final velocity, so we need another equation. Since we know that this is an elastic collision, energy must be conserved so we can also use:

$${{1} \over {2}} m_{2 balls}{v_{i}}^{2}= {{1} \over {2}} m_{f}{v_{f}}^{2}$$

If we substitute the $\LARGE{v_{f}}$ that we found in the momentum conservation equation into the energy equation we can find $\LARGE{m_{f}}$:

$${{1} \over {2}} m_{2 balls}{v_{i}}^{2}= {{1} \over {2}} m_{f}({m_{2 balls} \over m_{f}} v_{i})^2$$

Dividing out the parts that are the same on both sides and separating the squared part:

$$ m_{2 balls} {v_{i}}^{2} = m_{? balls} {{{m_{2 balls}}^{2}} \over {{m_{? balls}}^{2}}}{v_{i}}^{2}$$

Cancelling $ \LARGE{{v_i}^2} $ on both sides and $ \LARGE{m_{?balls}} $ on the right

$$m_{2 balls}={{{m_{2 balls}}^2} \over {m_{? balls}}}$$

Dividing both sides by $\LARGE{m_{2 balls}}$:

$$1={{m_{2 balls}} \over {m_{? balls}}} \therefore m_{2 balls}=m_{? balls}$$

Putting this back into the momentum equation we see that if the mass must be the same, then the velocity must also be the same:

$$ m_{2 balls} v_{i}= m_{2 balls}v_{f} \rightarrow v_{i}=v_{f}$$

This is always what happens with the cradle. Whenever a set number of balls is dropped on one side, the same number of balls go up on the other side to approximately the same height (which was shown to be directly related to velocity).

Back to Top

Presenting The Demo

This is a great demo to start with. Most people have seen it and it is very simple and easy to understand visually. The physics behind the demo section goes into much more detail than is necessary; the point of this demo is to explain simply, that momentum must be conserved. The best method is to explain, ask a question, then check with the demo. Be sure to keep your audience’s age in mind when explaining. For younger students there is no need to go into the math.

Each tab contains a different method for presenting this demo as written by a former outreach student.

This demonstration shows what [conservation of momentum] is. First off, does anyone know what momentum is (encourage any response, right or wrong, and try to use them to lead to the next part)? Momentum is an object’s mass multiplied by its velocity. That means the heavier something is or the faster it is moving, the more momentum it has. So which would have more momentum, a semi-truck going 50 miles an hour or a motorcycle going the same speed? The useful thing about momentum is that it must be conserved. Conservation in physics is a fancy way of saying that something’s value won’t change before and after an event.

What that means is that if two things crash into one another we can use their combined momentum before the crash to predict what will happen after the collision. So Newton’s cradle here can show us exactly that. When I drop one of these balls with a mass of say one ball, it will have a set speed at the bottom.

When it collides with the other balls (let the ball go) it will push exactly one ball up at the same speed on the other side.

Now that we know that, what will happen if I drop two balls together? (give them a chance to predict then show it). What about 3?

(For fun I like to go through all the balls up to and including just swinging all of them as technically it still works and the kids usually laugh at that.)

Not only does this show us that momentum and energy must be conserved, but also that we can use our understanding of physics to predict what will happen!

Additional topics that can be covered

Newton’s Cradle can also show the difference between elastic and inelastic collisions. This, however, is better shown with the Happy and Sad Ball demonstration.

Conservation of energy can be brought up, as well as conversion from potential to kinetic energy when lifting and letting the balls go.

Follow Up Demo

A good demonstration to follow up with would be the Large and Small Ball Collision demonstration. Newton’s cradle shows the simplest example of conservation of momentum and this demo makes it one step more complex by having different masses.

 

Tips

 

Back to Top

References

  1. Reference Missing!
  2. Reference Missing!

Back to Top