# collisions102

Demonstration In the Gym

Drop a basketball, from a known height,

observe the complete cycle of motion

measure the new height to which it bounces back,

Drop a tennis ball observe and measure.

Does the basketball return to the original height?

Does the tennis ball return to the original height?

Ooh, what happened ? Does the applied maths land Conservation of Momentum not now apply?

This might be worth investigating on your own,

Drop a Ball from a known height, measure the height to which it bounces back on its 1st 2nd and 3rd bounce.

Find the ratio of the height bounced back over the height it started at.

(the heights after bounces stand for the starting height of the subsequent ratios).

...... but before you run home and do that ........... do this...........

Drop a basketball and tennis ball simultaneously and the tennis ball on top

and measure what you can!

Homework

Calculate the momentum of each ball

mass of tennis ball = 57g = 0.057 kg

mass of basket ball = 600g = 0.6kg

the FIBA claim the mass must between 567 grams (g) and 650 g. They also pass as a more stringent requirement "it must bounce at least 1300 mm when dropped from a height of 1800 mm on a hard surface with a mass of more than 1 ton", I find this curious ... more later.

Of course if you don't believe me why don't you measure this yourself

The velocity of the balls can be calculated using uvast equations.

How could you prove this to be experimentally true?

And dont forget if you can

This might be worth investigating on your own,

Drop a Ball from a known height, measure the height to which it bounces back on its 1st 2nd and 3rd bounce.

Find the ratio of the height bounced back over the height it started at.

(the heights after bounces stand for the starting height of the subsequent ratios).

Impacts

this seems like an in life reality that we can simply simulate in maths,

but when striking a fixed body the momentum is conserved,

but we cannot measure the momentum imparted,

we can calculate the velocity from which it is rebounded back up.

The balls we dropped in the gym certainly didn't return back to where they came from. If you carried out the experiment at home and found the value of the ratio you should have found it to be less than 1 ... like 0.813 or so. (if you found it to be bigger than one, you got the ratio the wrong way around).

Collisions can be elastic or inelastic. For an elastic collision the momentum and energy are both conserved, thus an inelastic collision these values are not conserved. When we 'drop' a ball it does not return back to the height we dropped it from. This is what is meant be an inelastic collision. Such collisions are governed by the coefficient of restitution, e.

Neither the momentum nor the energy (of the measurable bodies) can be conserved due to the fact the velocities change with e

## The coefficient of Restitution e

If you carry out experiments (RYAN) then you will see that they bounce back a fixed factor of the height they fell, this fixed factor is called the coefficient of restitution and given the symbol e.

These balls lose energy in the bounce and the noise, this is reflected in the loss of velocity. The ratio of the velocities of the balls is the same as the ratio of the heights bounced back over the dropped height.

Newtons experimental law of Restitution (N.E.L)

u is the initial speed, v the velocity after the collision, _{1 & 2} represent the 2 bodies that are colliding.

So if the object is fixed, like the ground or a wall then it has no velocity before or after so the equation becomes

This is the 2nd equation that you will NEED to use in Every question, usually solving this leads to 5 marks.

http://vnatsci.ltu.edu/s_schneider/physlets/main/momenta3c.shtml

http://en.wikipedia.org/wiki/Coefficient_of_Restitution#Equation

This value *-e* is called the coefficient of Restitution , the minus sign represents the directional nature of the motion of the ball.

One measurement is going up and the other is going down, these velocities have opposite signs.

by using the V^{2} formula we can write the coefficient of non elastic collisions as ... go on you try it

Newtons experimental law of Restitution (N.E.L)

To find the return velocity we can rearrange the formula, or use equivalent fractions especially when the ration is expressed as a fraction, (surd).

Recall from the last section what is meant by impulse. It is the mass of the body times the change in velocity.

You have to remember that the velocities of a rebounding object is the addition of the two speeds, the difference between the velocities.

look at this applet BUT BE SURE TO PLAY WITH e

http://surendranath.tripod.com/Applets/Dynamics/Collisions/CollisionApplet.html

You should now go and try questions 7.B 1,2 & 3

Kinetic Energy is the energy a moving body has, this is related to its mass and to its velocity as per the equation

Therefore the changed speed of the rebounded object will have lead to a decrease in the kinetic energy of the object. We can find the change in energy by calculating the kinetic energy value before the collision and then again after the collision.

You should now complete those questions up to 5.

For the questions after 5 go on to collisions 103

Collisions 101 collisions102 collisions103 Collisions104 Collision105