Physics A Level
Chapter 6: Momentumا 6.5 Collisions in two dimensions
Physics A Level
Chapter 6: Momentumا 6.5 Collisions in two dimensions
It is rare that collisions happen in a straight line–in one dimension. Figure 6.14 shows a two-dimensional collision between two snooker balls. From the multiple images, we can see how the velocities of the two balls change:
At first, the white ball is moving straight forwards. When it hits the red ball, it moves off to the right.
Its speed decreases; we can see this because the images get closer together.
The red ball moves off to the left. It moves off at a bigger angle than the white ball, but more slowly – the images are even closer together.
How can we understand what happens in this collision, using the ideas of momentum and kinetic energy?
At first, only the white ball has momentum, and this is in the forward direction. During the collision, this momentum is shared between the two balls. We can see this because each has a component of velocity in the forward direction.
directions
At the same time, each ball gains momentum in the sideways direction, because each has a sideways component of velocity – the white ball to the right, and the red ball to the left. These must be equal in magnitude and opposite in direction, otherwise we would conclude that momentum had been created out of nothing. The red ball moves at a greater angle, but its velocity is less than that of the white ball, so that the component of its velocity at right angles to the original track is the same as the white ball’s.
Figure 6.15a shows the momentum of each ball before and after the collision. We can draw a vector triangle to represent the changes of momentum in this collision (Figure 6.15b). The two momentum vectors after the collision add up to equal the momentum of the white ball before the collision. The vectors form a closed triangle because momentum is conserved in this two-dimensional collision.
closed vector triangle shows that momentum is conserved in the collision
Components of momentum
Momentum is a vector quantity and so we can split it into components in order to solve problems.
Worked example 3 shows how to find an unknown velocity.
Worked example 4 shows how to demonstrate that momentum has been conserved in a two-dimensional collision.
WORKED EXAMPLES
3) A white ball of mass $m = 1.0 kg$ and moving with initial speed $u = 0.5\,m\,{s^{ - 1}}$ collides with a stationary red ball of the same mass. They move off so that each has the same speed and the anglev between their paths is ${90^ \circ }$. What is their speed?
Step 1: Draw a diagram to show the velocity vectors of the two balls, before and after the collision (Figure 6.16). We will show the white ball initially travelling along the y-direction.
Because we know that the two balls have the same final speed v, their paths must be symmetrical about the y-direction. Since their paths are at ${90^ \circ }$ to one other, each must be at ${45^ \circ }$ to the y-direction.
Step 2: We know that momentum is conserved in the y-direction. Hence we can say:
initial momentum of white ball in y-direction
= final component of momentum of white ball in y-direction
+ final component of momentum of red ball in y-direction
This is easier to understand using symbols:
$mu = mvy + mvy$
where vy is the component of v in the y-direction. The right-hand side of this equation has two identical terms, one for the white ball and one for the red. We can simplify the equation to give:
$mu = 2m{v_y}$
Step 3: The component of v in the y-direction is v cos ${45^ \circ }$. Substituting this, and including values of m and u, gives
$0.5 = 2v\,\cos {45^ \circ }$
and hence
$v = \frac{{0.5}}{{2\,\cos {{45}^ \circ }}} \approx 0.345\,m\,{s^{ - 1}}$
So each ball moves off at $0.345\,m\,{s^{ - 1}}$ at an angle of ${45^ \circ }$ to the initial direction of the white ball.
4) Figure 6.17 shows the momentum vectors for particles 1 and 2, before and after a collision. Show that momentum is conserved in this collision.
Step 1: Consider momentum changes in the y-direction.
Before collision:
$momentum = 0$
(because particle 1 is moving in the x-direction and particle 2 is stationary).
After collision:
component of momentum of particle 1
$ = 3.0\,\cos \,{36.9^ \circ }\, \approx \,2.40\,kg\,m\,{s^{ - 1}}\,upwards$
component of momentum of particle 2
$ = 4.0\,\cos \,{53.1^ \circ }\, \approx \,2.40\,kg\,m\,{s^{ - 1}}\,upwards$
These components are equal and opposite, and hence their sum is zero. Hence, momentum is conserved in the y-direction.
Step 2: Consider momentum changes in the x-direction.
Before collision:
$moment = 5.0\,kg\,m\,{s^{ - 1}}$ to the right
After collision:
component of momentum of particle 1
$ = 3.0\,\cos \,{53.1^ \circ }\, \approx \,1.80\,kg\,m\,{s^{ - 1}}$ to the right
component of momentum of particle 2
$ = 4.0\,\cos \,{36.9^ \circ }\, \approx \,3.20\,kg\,m\,{s^{ - 1}}$ to the right
total momentum to the right = 5.0 kg m s−1
Hence, momentum is conserved in the x-direction.
Step 3: An alternative approach would be to draw a vector triangle similar to Figure 6.15b. In this case, the numbers have been chosen to make this easy; the vectors form a 3–4–5 rightangled triangle.
Because the vectors form a closed triangle, we can conclude that:
momentum before collision = momentum after collision (in other words, momentum is conserved)
Questions
9) A snooker ball strikes a stationary ball. The second ball moves off sideways at ${60^ \circ }$ to the initial path of the first ball.
Use the idea of conservation of momentum to explain why the first ball cannot travel in its initial direction after the collision. Illustrate your answer with a diagram.
10) Look back to Worked example 4. Draw the vector triangle that shows that momentum is conserved in the collision described in the question. Show the value of each angle in the triangle.
11) Figure 6.18 shows the momentum vectors for two identical particles, 1 and 2, before and after a collision. Particle 2 was at rest before the collision. Show that momentum is conserved in this collision.
12) A snooker ball collides with a second identical ball as shown in Figure 6.19.
a: Determine the components of the velocity of the first ball in the x- and y-directions.
b: Hence, determine the components of the velocity of the second ball in the x- and y-directions.
c: Hence, determine the velocity (magnitude and direction) of the second ball.
