Physics Puzzle: Newton’s Cradle
We’ve all seen this executive toy somewhere:
As the ball comes down and a collision occurs, it comes to rest while another ball at the other end rises up. This repeats ad infinitum - at least in theory - and illustrates perfectly the principle of conservation of momentum - the momentum of the colliding ball on one end is “transferred” to the ball at the other end, because it can’t just “go away”.
Let’s imagine a different scenario. Let all of our balls be of identical size and shape, each having mass m. Now suppose that the ball coming down (in the photo above) is of mass 2m instead of m. Is seems that conservation of momentum would dictate that this time, TWO balls should rise up on the other end.
Is this correct, or is there some hidden fallacy in the argument?
The 2 balls at the end are not glued together; in fact, none of the balls are. Hence, we should think of the balls as infinitesimally separated from each other, and after the ball comes down on the left, we have separate collisions between it and the first ball, then the first ball and the second, second and the third, and so forth. This follows from the assumptions that we have contact forces between the balls, and that Newton’s 3rd law holds. By conservation of momentum, this tells us the last ball on the right should start moving to the right at the speed twice that of the ball coming down on the left.
Good try. I’ll withhold my judgment for the time being, and remark that according to your analysis it seems to me that no matter what hits the balls on one side, a single ball will rise up on the other side.
However, we all know that if we stretch back and release TWO balls in a simple Newton’s cradle (all masses equal), then two balls will also rise up on the other end. How does your analysis explain this?
If you release 2 balls on the left, my guess is, what happens is the second ball closer to the rest of the balls hits first. This causes the rightmost ball to go flying. But within a split second the leftmost ball hits the second ball from the left, and sets off yet another ball flying — hence two balls on the right ends up rising together.
Momentum is ‘The product of the mass times the velocity of an object.’ So, there can be two cases:-
Case A:
One ball rises up but has twice the velocity.
Case B:
Two balls rise up (to contribute the mass of 2m) but velocity remains same as the initial ball released.
I’ll go for case A. I assume its the right answer. Plz let me know.
The balls’ collisions are almost elastic, and hence you need not to merely conserve momentum, but also conserve energy. If a single ball would rise up with twice the velocity, this would not be the case, since:
Energy of “big” ball - (2m)v^2/2
Energy of “small” ball - m(2v)^2/2
which are not equal. On the other hand, the second scenario, B, fits the bill perfectly, and is therefore the correct solution, since:
Energy of two small balls - mv^2/2 + mv^2/2
Which coincides with that of the initial ball.
Well, if two balls are lifted and dropped, two ball bounce away. Similarly, if a ball of twice the mass 2m hits the left side, two balls of mass m will bounce away, in accordance with the conservation of momentum.
Jac.