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## Tiger Woods and his driver

In a previous blog I mentioned that I would derive some equations relating to two body collisions, which would, among other things, allow one to see how quickly (and hence how far) one can hit a golf ball with a golf driver. I do this in this blog. I am sorry for it being so long, and also so mathematical, but there is really no other way to derive these important relationships than to go through the mathematics.

## Conservation of momentum and elastic collisions

In all collisions, momentum, defined as $\vec{p} = m\vec{v}$ where $m$ is the mass and $\vec{v}$ is the velocity, is conserved. Momentum is a vector, so has both magnitude (size) and direction.

In an elastic collision, kinetic energy is also conserved. Kinetic energy is the energy of motion, and is defined as $KE=\frac{1}{2}mv^{2}$. We will consider two masses colliding, the mass of object 1 is $m_{1}$, the mass of object 2 is $m_{2}$. The velocity before the collision of object 1 is $u_{1}$, and of object 2 is $u_{2}$, the velocity after the collision of object 1 is $v_{1}$ and that of object 2 is $v_{2}$.

The conservation of momentum allows us to write $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$ and the conservation of kinetic energy allows us to write $\frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m_{2}u_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$

## Collisions where object 2 is stationary before the collision

In this blog, I am going to consider the case of object 2 having no velocity before the collision. This is, in fact, very common. For example, it is the sort of collision we have when Tiger Woods’ golf driver hits his golf ball as he drives it up the fairway (or when his ex-wife allegedly hit his car with a golf club when she learned of his affairs!). Tiger Woods driving his golf ball up the fairway is an example of a two body collision

## The expression for the velocity of object 1 $v_{1}$ after the collision

If object 2 is stationary before the collision then $u_{2}=0$ and so we can write, for the conservation of momentum, $m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2} \qquad(1)$ and, for conservation of kinetic energy, $\frac{1}{2}m_{1}u_{1}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} \qquad(2)$.

Re-arranging equation (1) we get $m_{2}v_{2} = m_{1}u_{1} - m_{1}v_{1} = m_{1}(u_{1}-v_{1})$ and so $v_{2} = \frac{m_{1}(u_{1}-v_{1})}{m_{2}} \qquad(3)$.

Multiplying each term in equation (2) by 2, and re-arranging, we get $m_{2}v_{2}^{2} = m_{1}u_{1}^{2} - m_{1}v_{1}^{2} = m_{1}(u_{1}^{2} - v_{1}^{2})$ and so $v_{2}^{2} = \frac{m_{1}(u_{1}^{2} - v_{1}^{2})}{m_{2}} = \frac{m_{1}(u_{1}+v_{1})(u_{1}-v_{1})}{m_{2}} \qquad(4)$

If we now square equation (3) we get $v_{2}^{2} = \frac{m_{1}^{2}(u_{1}-v_{1})(u_{1}-v_{1})}{m_{2}^{2}} \qquad (5)$

We can set this equal to equation (4), which allows us to write $\frac{m_{1}^{2}(u_{1}-v_{1})(u_{1}-v_{1})}{m_{2}^{2}} = \frac{m_{1}(u_{1}+v_{1})(u_{1}-v_{1})}{m_{2}}$. Cancelling common terms on both sides, this becomes $\frac{m_{1}(u_{1}-v_{1})}{m_{2}} = u_{1}+v_{1}$ which can then be re-arranged to give $m_{1}(u_{1}-v_{1}) = m_{2}(u_{1}+v_{1})$. Re-arranging this to get the terms involving $v_{1}$ all on one side, $m_{1}v_{1}+m_{2}v_{1} = m_{1}u_{1}-m_{2}u_{1}$ and so we can write $v_{1}(m_{1}+m_{2}) = u_{1}(m_{1}-m_{2})$.

So, finally, the expression for the velocity of object 1 after the collision is $\boxed { v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} }$

## The expression for the velocity of object 2 after the collision, $v_{2}$

To find $v_{2}$ we need to substitute this value for $v_{1}$ from equation (3), so we can write $m_{2} v_{2} = m_{1} u_{1} - m_{1}v_{1} = m_{1} u_{1} - m_{1} \left ( \frac{u_{1} (m_{1} -m_{2})} {(m_{1} + m_{2}) } \right )$
which can then be written $m_{2} v_{2} = \frac{ m_{1} u_{1} ( m_{1} + m_{2} ) - m_{1} u_{1} ( m_{1} - m_{2} ) } { m_{1} + m_{2} }$

Which becomes $m_{1} v_{2} ( m_{1} + m_{2} ) = m_{1}^{2} u_{1} + m_{1} m_{2} u_{1} - m_{1}^{2} u_{1} + m_{1} m_{2} u_{1}$

which becomes $v_{2} ( m_{1} + m_{2} ) = 2 m_{1} u_{1}$

So, finally, the expression for the velocity of object 2 after the collision is $\boxed{ v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } }$

## The case when $m_{1}<;<;m_{2}$

When a moving object strikes a stationary object which is much more massive, what will happen? Intuition tells us that the stationary object will not move, and that the incoming moving object will rebound. For example, this is what happens when a ball hits a wall.

The equations for $v_{1}$ and $v_{2}$ which we derived above can show us that this is indeed true. If we make $m_{1} <;<; m_{2}$ in the equation for $v_{1}$ we get $v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} \approx \frac{u_{1}(-m_{2})}{m_{2}} \approx -u_{1}$. Remembering that, in fact, $v_{1}$ and $u_{1}$ are vectors, so in fact we should write $\boxed{\vec{v_{1}} \approx -\vec{u_{1}}}$, so the velocity after the collision with a stationary, more massive object, is that the object returns in the opposite direction to its initial direction, but with the same speed.

When an object strikes a stationary object, we can also work out the conditions necessary for the incoming object to rebound. In other words, the conditions for $v_{1}$ to be negative. One can see that all that is required for $v_{1}$ to be negative is that $m_{2} >; m_{1}$. So, whenever a less massive object strikes a stationary more massive object, the incoming less massive object will rebound.

In the famous Geiger-Marsden experiment, performed at the University of Manchester in 1909 by Hans Geiger and Ernest Marsden, under the direction of Ernest Rutherford, they found that some alpha-particles fired at a thin sheet of gold foil were bouncing back towards the source. It is because of the equation above for $v_{1}$ that Rutherford knew the alpha-particles must be striking something more massive, which he correctly concluded was a densely packed nucleus in the gold atoms.

## The case when $m_{1} = m_{2}$

What about when the mass of the two objects is the same? In this case, we can get an idea from Newton’s cradle, as shown in my blog of a few weeks ago, and illustrated in this video when one ball hits the other 4 stationary balls.

What happens is that the incoming ball transfers all its momentum to the stationary ball, so that the first ball stops and the second ball moves in the same direction as the original ball, and with the same speed. If we let $m_{1}=m_{2}=m$ then, mathematically, we can see that, for $v_{1}$ $v_{1}$ we get $v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} = \frac{u_{1}(m-m)}{(m+m)} = 0$

so object 1 does, indeed, come to rest after the collision. But, for object 2 we can write $v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } = \frac{ 2m u_{1} }{2m} = u_{1}$

so object 2 moves off after the collision with the same velocity that object 1 had before the collision. This can all be seen in the video above.

## The case when $m_{1} >;>; m_{2}$

What about when the mass of the incoming object is much greater than the mass of the stationary object? In this case, the equation for $v_{1}$ becomes $v_{1} = \frac{u_{1}(m_{1}-m_{2})}{(m_{1}+m_{2})} \approx \frac{u_{1}m_{1}}{m_{1}} \approx u_{1}$

so the 1st object carries on after the collision with the same velocity it had before the collision, as if the less massive object were not there.

For the initially stationary less massive object, we can calculate $v_{2} = \frac{ 2 m_{1} u_{1} }{ (m_{1} + m_{2} ) } \approx \frac{2m_{1}u_{1}}{m_{1}} \approx 2u_{1}$

which may come as a surprise to you. The initially stationary object will move off after the collision with twice the velocity of the incoming more massive object.

## Applying these equations to golf

When Tiger Woods is trying to drive his golf ball up the fairway, he is of course not using a club head which is infinite in mass. The typical mass of a golf ball is probably about 40 grams, with the maximum allowed being 45.9 grams. The mass of the head of a golf club seems to be less regulated, I cannot find any regulations governing the mass of a golf club head, although I did find some regulations about the volume. But, I did find that the typical mass of a golf driver head is about 200g.

So, for simplicity, less us assume the mass of the golf driver head is 4 times the mass of the golf ball, which is approximately true. That is, $m_{1}=4m_{2}$. Then, we can show that the maximum value for $v_{2}$, the velocity of the golf ball after it has been hit, is $v_{2} = \frac{2m_{1}u_{1}}{m_{1}+m_{2}} = \frac{8m_{2}u_{1}}{5m_{2}} = \frac{8u_{1}}{5}$

so the maximum velocity the golf ball can have (and hence how far it can go) is $1.6 u_{1}$, the velocity of the golf club head. The faster one can swing the club, the further the ball will go, but with these figures where $m_{1} = 4m_{2}$, it can never acquire a velocity which is more than 1.6 times the velocity of the golf club head.

In a (much shorter) future blog, I will explain why it would not necessarily help to increase the golf club head mass, even though in theory one could get close to having $v_{2}$ to be $2u_{1}$ rather than the $1.6 u_{1}$ we get here.

### One Response

1. […] Bohr suggested in a paper in 1913 that electrons would somehow not radiate away their energy if they were orbiting in certain “allowed orbits”. If they were in these special orbits, the normal laws of EM radiation would not apply. He suggested that these allowed orbits were when the orbital angular momentum could be written as ( is Planck’s constant, and is given its own symbol in Physics at it crops up so often). What is orbital angular momentum? Well, it is the rotational equivalent of linear momentum. Linear momentum is defined as . Notice, momentum is a vector quantity, this is important in doing calculations involving collisions, such as the ones I did in this blog. […]