Extra force in other frame paradox

Question

Suppose there are two balls: one steel ball that cannot break with a mass of 10 kg, and another smaller wooden ball with a mass of 1 kg, which will break if a force of 10 N is applied to it.

Consider a scenario where the wooden ball is moving toward the stationary steel ball with a velocity of 1 m/s. The collision between the two balls is completely inelastic, meaning that they stick together after the collision.

In the frame of reference attached to the steel ball, the wooden ball has a momentum of 1 kgm/s (since its mass is 1 kg and its velocity is 1 m/s). When the collision occurs, the wooden ball will come to a complete stop or move slowly together(lets say that v_f=0) , which suggests a change in momentum of 1 kgm/s. The force experienced by the wooden ball would be $ \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum. If this force is calculated to be 1 N, it would not be enough to break the wooden ball.

Now, let's look at this from the frame attached to the wooden ball. In this frame, the steel ball is moving toward the wooden ball with a velocity of 1 m/s, carrying a momentum of 10 kg*m/s (since its mass is 10 kg and its velocity is 1 m/s). Upon collision, the steel ball will come to a stop as seen from the wooden ball. This suggests a much larger change in momentum, which would imply a force of 10 N, enough to break the wooden ball.

This appears paradoxical because, from the steel ball's frame, the wooden ball should not break, while from the wooden ball's frame, it should break due to the larger force. How can this paradox be resolved?

I know its answer is somewhere when the collision occurs the wooden ball's frame becomes noninertial but how does that wooden ball know that I am currently noninertial since in its own frame its always at rest and the force that is going to act upon me is due to my own declaration.

Answer 1

Assuming you want to work with inertial frames, to avoid issues of fictitious forces...

There is no one frame of the wooden ball, because its velocity changes before and after the collision. So we should talk about the frame at which the wooden ball is initially at rest.

In this frame, the steel ball does not come to rest. Rather, after the collision, it will continue at approximately the same speed (but a little less, so that momentum is conserved). Instead, the wooden ball will speed up, to match the speed of the steel ball.

The force on the wooden ball will be determined by the change in momentum of the wooden ball. Since the mass is $1 {\rm kg}$ and the final velocity will be about $1 m/s$, then assuming $\Delta t\approx 1 s$, then $\frac{\Delta p}{\Delta t} \approx 1 N$. This is as it should be. The net force on an object must the same in any inertial reference frame, because in Newton's second law $\sum F = m a$, the right hand side does not depend on the reference frame, so neither can the left hand side.

As an aside, $\Delta t \approx 1 s$ is probably way too big for a realistic collision. Normally in introductory mechanics classes, we don't worry too much about $\Delta t$ because it will depend on a lot of complicated factors like the elasticity of the two objects, and we basically assume that it is zero or much smaller than any other time scale in the problem. As a result, the instantaneous force is not something we usually need to know. You can apply a huge force to an object over a very tiny time interval, and not break it. Typically the amount of momentum and energy transferred to the object is more important than the instantaneous force in a collision. This can be contrasted with the situation of a constant force being applied over a long time, which leads to stress-strain curves where the object deforms and can ultimately break under the stress of an applied force.

Answer 2

In such collision problems, you usually do not look at the forces during the collision. You look at velocities, momenta, and or energies before and after. These are usually problems about conservation laws.

Think of a soft, springy rubber ball bouncing off the floor. It flattens as it hits the floor. Parts of it stretch. Kinetic energy turns into potential energy. The forces needed to stretch the ball slow it to a stop. The stretched parts spring back. The forces throw the ball off the floor, converting potential energy into kinetic energy.

Harder balls like steel and wood are harder to deform. They may deform very little. If so, the collision takes place in the very short time they are in contact and forces are very large. You may find that forces are so large that a ball you think of as rigid deforms significantly.

Titleist Golf Balls at The Moment of Impact

It takes work to calculate what happens during such a collision. You cannot treat the ball as a single rigid object. Different parts are traveling at different speeds, and are subject to different forces.

You can track what happens to each part from any inertial frame of reference. You will find the velocities are different, but the accelerations and forces are the same.

You will find energy and momentum are conserved during the collision. You have the same energy and momentum before and after. And of course, you can treat the balls as simple rigid objects before and after.