3
$\begingroup$

When riding a bike on a bumpy road, objects (let's say, a baseball weighing 150 grams) sometimes fly upwards and out of the basket.

However I cannot seem to be able to explain this physically. Instead, I get:

  1. The upwards speed of the baseball will be equal to upwards speed of the basket.

  2. The bike and the baseball will fall at the same acceleration.

  3. Therefore distance between baseball and basket will remain 0.

I have so far two possible explanations:

A. There is an elastic collision between the bike and the baseball. The baseball is lighter and thus receives larger upward speed.

B. The impulse travels through the bike similarly to how it does in a Newton's cradle.

I think that both of these phenomenon are occurring there, and perhaps something else also. What is the relative importance of these, i.e. what really causes the objects to fly up?

||cite|improve this question
$\endgroup$
  • $\begingroup$ Newton's cradle is just an example of elastic collisions in which the masses are equal. $\endgroup$ – Brionius Dec 1 '15 at 15:21
  • $\begingroup$ What's the basic difference between the basket and the ball? (hint: one is tied to the bicycle) $\endgroup$ – Carl Witthoft Dec 1 '15 at 15:24
  • 1
    $\begingroup$ Not sure why this question got a downvote. I think it's quite interesting. $\endgroup$ – Floris Dec 1 '15 at 15:25
  • $\begingroup$ Carl - just say "the whole bike" then. $\endgroup$ – Fattie Dec 1 '15 at 15:37
  • 2
    $\begingroup$ Floris -- "a little knowledge is a dangerous thing". :) Superficially the question seems simplistic. But it's actually a great engineering question of course, as you say. $\endgroup$ – Fattie Dec 1 '15 at 15:37
4
$\begingroup$

When two objects collide, they transfer momentum because they exert an equal and opposit force on each other (Newton's third law), and $\Delta p = \int F dt$.

In order to know how fast an object moves after a collision, we need to know the velocities and mass of the objects before the collision and how elastic the collision is (conservation of kinetic energy). For now, we will assume a perfectly elastic collision between ball and bicycle basket. We will also assume the basket is firmly attached to the bike, and that the mass "bike plus basket" is much greater than that of the ball.

Now we can analyze the situation: it is like that of a ball hitting a moving wall. If a ball moves to the right at velocity $v_1$ and hits a wall moving to the left at velocity $v_2$, then their relative velocity is $v_1 + v_2$. Now comes a trick: we look at the collision from the point of view of the wall (really we want to use the "center of mass frame", but since the wall is much heavier than the ball, we can use the wall instead). The wall sees the ball come towards it at $v_1+v_2$, and after an elastic collision it needs to move away at the same speed $v_1+v_2$. This means the ball velocity changed by $2(v_1+v_2)$. That same change must be observed in the "world" frame of reference, so after a collision the velocity is

$$v' = 2(v_1+v_2) - v_1= v_1 + 2 v_2$$

to the left.

In other words - the ball picks up twice the velocity of the thing that is hitting it.

And that is fundamentally what happens for you. The ball bounces around, and if you are unlucky it is moving down just as the basket moves up. The sum of their velocities can be sufficient to throw the ball out of the basket: and if it wasn't, then the ball's $v_1$ will be greater on the next bounce.

Elastic collisions, one mass much greater than the other, and random inputs of velocity from the basket. Those three ingredients are sufficient to explain this.

Afterthought: it is more likely for the ball to hit the basket while their relative velocities are greatest: if you consider the motion of the basket to be truly random, then during the time interval that it is moving away from the ball the rate at which the ball approaches is low: when the basket moves towards the ball the rate is high. So a randomly moving basket is more likely to give additional energy to the ball rather than remove it! In a way it is "designed" to throw the ball out.

||cite|improve this answer
$\endgroup$
  • $\begingroup$ Seems reasonable. I forgot to take into account the amplifying effect of multiple bounces. $\endgroup$ – jpa Dec 1 '15 at 16:30
  • $\begingroup$ So, if your logic holds true, then a baseball that is thrown at the same average velocity as a baseball bat that is swung at the same average velocity means the ball will travel at 3 times the average velocity (assuming a perfectly elastic collision) after they collide? $\endgroup$ – DIYser Dec 1 '15 at 16:53
  • $\begingroup$ @DIYser if the bat is much heavier than the ball that would be true. Otherwise you have to use the velocity of the center of mass, not the velocity of the bat. This will reduce the factor from 1+2 to $1+\frac{2M}{m+M}$ $\endgroup$ – Floris Dec 1 '15 at 17:12
  • $\begingroup$ I also think this might be made worse because a bike is not a truly rigid structure. It can store energy by distorting the frame when a bump gets hit, and release it like a spring, sending the baseball upwards while the CM of the bike itself moves far less. $\endgroup$ – Cort Ammon Dec 1 '15 at 18:08
  • $\begingroup$ @CortAmmon maybe - but a basket itself is very light so if it is loose the transfer of momentum will be less (except for things like ping pong balls). The "stored energy" is mostly just a statement of the elastic nature of the collision unless you get a slap shot effect (basket moving faster than bike like the stick moves faster than the hands in a slap shot). Whether the additional velocity offsets the lower mass is something I could only guess. $\endgroup$ – Floris Dec 1 '15 at 18:16
-1
$\begingroup$

The force upon both objects is equal, but since the mass of the baseball is less, the acceleration will be greater due to F=MA

||cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.