Springy collisions

Figure 6.3a shows what happens when one snooker ball collides head-on with a second, stationary ball.
- The result can seem surprising. The moving ball stops dead. The ball initially at rest moves off with the same velocity as that of the original ball. To achieve this, a snooker player must observe two conditions:
- The collision must be head-on. (If one ball strikes a glancing blow on the side of the other, they will both move off at different angles.)
The moving ball must not be given any spin. (Spin is an added complication that we will ignore in our present study, although it plays a vital part in the games of pool and snooker.)
You can mimic the collision of two snooker balls in the laboratory using two identical trolleys, as shown in Figure 6.3b. The moving trolley has its spring-load released, so that the collision is springy. As one trolley runs into the other, the spring is at first compressed, and then it pushes out again to set the second trolley moving. The first trolley comes to a complete halt. The ‘motion’ of one trolley has been transferred to the other.
You can see another interesting result if two moving identical trolleys collide head-on. If the collision is springy, both trolleys bounce backwards. If a fast-moving trolley collides with a slower one, the fast trolley bounces back at the speed of the slow one, and the slow one bounces back at the speed of the fast one. In this collision, it is as if the velocities of the trolleys have been swapped.

Sticky collisions

Figure 6.4 shows another type of collision. In this case, the trolleys have adhesive pads so that they stick together when they collide. A sticky collision like this is the opposite of a springy collision like the ones described previously.

If a single moving trolley collides with an identical stationary one, they both move off together. After the collision, the speed of the combined trolleys is half that of the original trolley. It is as if the ‘motion’ of the original trolley has been shared between the two. If a single moving trolley collides with a stationary double trolley (twice the mass), they move off with one-third of the original velocity.
From these examples of sticky collisions, you can see that, when the mass of the trolley increases as a result of a collision, its velocity decreases. Doubling the mass halves the velocity, and so on.

Defining linear momentum

From the examples discussed earlier, we can see that two quantities are important in understanding collisions:
- the mass m of the object
- the velocity v of the object.
These are combined to give a single quantity, called the linear momentum (or simply momentum) p of an object.

The momentum of an object is defined as the product of the mass of the object and its velocity. Hence:

$\begin{array}{l}
momentum\, = \,mass\, \times \,velocity\\
p\, = \,mv
\end{array}$

The SI unit of momentum is $kg\,m\,{s^{ - 1}}$. There is no special name for this unit in the SI system. The newton second (N s) can also be used as a unit of momentum (see topic 6.7).
Momentum is a vector quantity because it is a product of a vector quantity (velocity) and a scalar quantity (mass). Momentum has both magnitude and direction. Its direction is the same as the direction of the object’s velocity.
In the earlier examples, we described how the ‘motion’ of one trolley appeared to be transferred to a second trolley, or shared with it. It is more correct to say that it is the trolley’s momentum that is transferred or shared. (More precisely, we should refer to linear momentum, because there is another quantity called angular momentum that is possessed by spinning objects.)
As with energy, we find that momentum is also conserved. We have to consider objects that form a closed system–that is, no resultant external force acts on them. The principle of conservation of momentum states that, within a closed system, the total momentum in any direction is constant.
The principle of conservation of momentum can also be expressed as follows:
For a closed system where no resultant external force acts, in any direction:
total momentum of objects before collision = total momentum of objects after collision A group of colliding objects always has as much momentum after the collision as it had before the collision. This principle is illustrated in Worked example 1.

WORKED EXAMPLE

In Figure 6.5, trolley A of mass $0.80 kg$ travelling at a velocity of $3.0\,m\,{s^{ - 1}}$ collides head-on with a stationary trolley B. Trolley B has twice the mass of trolley A. The trolleys stick together and have a common velocity of $1.0\,m\,{s^{ - 1}}$ after the collision. Show that momentum is conserved in this collision.

Step 1: Make a sketch using the information given in the question. Notice that we need two diagrams to show the situations, one before and one after the collision. Similarly, we need two calculations – one for the momentum of the trolleys before the collision and one for their momentum after the collision.
Step 2: Calculate the momentum before the collision:
$\begin{array}{l}
momentum\,of\,trolleys\,before\,collision\\
 = {m_A}\, \times \,{u_A}\, + \,{m_B}\, \times \,{u_B}\\
 = (0.80\, \times \,3.0) + 0\\
 = 2.4\,kg\,m\,{s^{ - 1}}
\end{array}$
Trolley B has no momentum before the collision, because it is not moving.
Step 3: Calculate the momentum after the collision:
$\begin{array}{l}
momentum\,of\,trolleys\,before\,collision\\
 = ({m_A} + \,{m_B})\, \times \,{v_{A + B}}\\
 = (0.80\, + \,1.60) \times 1.0\\
 = 2.4\,kg\,m\,{s^{ - 1}}
\end{array}$
So, both before and after the collision, the trolleys have a combined momentum of $2.4\,kg\,m\,{s^{ - 1}}$. Momentum has been conserved.

Questions

 

2) Calculate the momentum of each of the following objects:
a: a $0.50 kg$ stone travelling at a velocity of $20\,\,m\,{s^{ - 1}}$
b: a 25 000 kg bus travelling at $20\,\,m\,{s^{ - 1}}$ on a road
c: an electron travelling at $2.7 \times {10^7}\,\,m\,{s^{ - 1}}$.
(The mass of the electron is $9.1 \times {10^{ - 31}}\,\,kg$.)

3) Two balls, each of mass $0.50 kg$, collide as shown in Figure 6.6. Show that their total momentum before the collision is equal to their total momentum after the collision.