# Changing Directions on a Dime!

### Overview

Have you ever played (or watched someone play) tennis or baseball? These sports all have one thing in common: they involve hitting a ball. In the case of baseball, the pitcher throws a ball toward the batter, and the batter tries to hit the ball as far as possible. Or in tennis the ball is hit back and forth between players using racquets.

So what is the purpose of a racquet or a bat? They both are used to impact the ball and change its direction. When the impact between the bat or racquet and the ball occurs, the ball sees an unbalanced force:

The result of an unbalanced force is a change in velocity. In fact the change is so great that it results in a change in direction! If the bat were not there the ball would continue to the catcher… no force from the bat, no change in direction! Whereas the ball was moving towards the batter, after it is struck it moves in the opposite direction (and hopefully is a homerun!). The effect of a tennis racquet is exactly the same as our bat example.

It is also important to note that the bat is extremely hard. Why? When the bat applies the force of the bat on the ball, the ball also exerts an equal and opposite force on the bat. This is Newton's second law in action! Practically, it means that the bat and the ball have to be hard enough to withstand such large forces. If you don't believe this, go outside and have someone pitch you a potato. Watch it explode when you hit it with the bat… it isn't hard enough to withstand the force!

Let's explore this effect the Chaos way!

#### Setup

Set up one piece of track, slightly slanted, about a foot above the Chaos trampoline.

### Activity

#### Stage 1

Start the ball at the top of the track and let it drop onto the trampoline. Observe how the ball changes direction as it impacts with the drum of the trampoline! It is just like when a ball hits a bat or a tennis racquet.

After the ball bounces up, note how high it bounces. Does it bounce up as high as the track? Why do you think it doesn't reach that height?

#### Stage 2: Energy Losses

Conservation of energy seems to say that it should bounce as high as it was dropped. Let's try to figure out why it isn't doing that!

#### Vibration

When the ball drops, the support structure moves a little bit. This is because it is not completely rigid. Have you ever noticed how the structure can shift a little while you assemble it? Well it can move little bits during operation too, which leads to energy losses.

Heat losses

When the drum expands and contracts during contact, the surface heats up (although an imperceptibly small amount). This is a source of energy loss.

Sound losses

When you drop the ball, can you hear anything when it hits the drum? What does the impact sound like?

Making the structure vibrate takes energy away from the motion of the ball. Likewise, the energy to make sound sucks some energy from the ball's motion. These combine to decrease the energy available for motion, and the ball does not bounce as high as it was dropped.

#### Stage 3: Quantitative Approach to Problems

Estimate the force exerted by the drum in reversing the direction of the ball. Impact is defined by:

F*delta (t) = m*delta (v)
F=m delta(v)/delta(t)

To use this formula to estimate the force exerted by the drum on the ball, we need to estimate the time that the ball is in contact with the drum. This is delta(t). Take a stopwatch and make a guess as to how long it is in contact with the drum. It doesn't have to be specific! Is it 1 second? 1/2 second? 1/4 second?

We also need to estimate delta(v). Let's do this by assuming that there are no sound or heat losses (even though we know that there are). If this is the case, then conservation of energy would say that:

mgh = 1/2mv²

therefore:

½
v = (2gh)

From this statement we can estimate the velocity of the ball just before impact with the drum by knowing the height from which it was dropped! Assuming that there are no energy losses from heat or sound implies not only that this is the velocity of impact, but it is also the velocity of exit, only in the opposite direction!

From this we see that the change in velocity is approximately:

delta(v) = v - (-v) = 2v

Therefore we now need to know the weight of the ball! If you have a scale, measure it! Otherwise, assume that it weighs 1 gram=0.001 kg.

Now calculate an estimate of the force exerted by the drum on the ball.

### Quiz Yourself

#### Stage 1

Set up the track to use the trampoline to bounce the ball into the basket. Based on your experiment, having watched how high it can bounce off the trampoline, try to put the basket at the highest point possible. What should you do if you want to get the ball to bounce higher?

[Drop it from a higher point!]

#### Stage 2

Vibration is a major source of energy loss in the Chaos toy. Test this by setting up the trampoline on the toy and see how high the ball bounces if you don't hold on to the toy. Now grab on to the base with your two hands to stabilize the toy while you drop the ball. Watch how the ball bounces higher!

#### Stage 3

A 3 kg ball is dropped from 1 meter above the drum. Assume the impact take place in 0.25 seconds. What would the force approximately from the drum on the ball? From the ball on the drum?

Solution

At impact we have an approximate velocity of:

½
v = (2gh)

½
v = (2*9.8*1)

v = 4.43m/s

The change in velocity is approximated by:

delta(v) = v - (-v) = 2v

delta(v) = 2 (4.43m/s) = 8.86m/s

The force of the ball on the drum is identical in magnitude, but opposite in direction to the force of the drum on the ball (Newton's law). The magnitude of the force is:

F= m delta(v)/delta(t)

F= (3kg)*(8.86m/sec)/(0.25sec)

F= 6.645 Newtons.