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4.5Two-dimensional collisions
Figure 4.22 shows a two-dimensional collision. The red ball collides with a stationary blue ball not exactly head on (we call this a glancing collision). As a result, the two balls move on the two-dimensional plane of the paper. We will mostly deal with situations in which one of the colliding particles is initially at rest.
The blue ball moves off with speed v in a direction making an angle Î¸ with the x axis. The x axis was chosen to be the initial direction of motion of the red ball. The y axis is at right angles to the x axis as usual.

Figure 4.22: A two-dimensional collision.

Our objective is to find v and Î¸.
We will now find components of momentum before and after the collision along the x and y directions and demand that the total momentum in each direction is the same as that after the collision, i.e.
âpyîîîîîîbeforeÂ =âpyîîîîîîafterÂ 
and similarly for the y direction,
âpyîîîîîîbeforeÂ =âpyîîîîîîafterÂ 
Applying these equations to our collision we find
xÂ direction:Â :2.0Ã6.0+0îîîîîîîîîîîîîîtotal beforeÂ =2.0Ã4.0Ãcos20Â°+4.0ÃvÃcosÎ¸îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotal afterÂ 
yÂ direction:Â 0îîîîîîîîîîîîtotal beforeÂ =2.0Ã4.0Ãsin20Â°â4.0ÃvÃsinÎ¸îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotal afterÂ 
These imply
7.52+4.0ÃvÃcosÎ¸=12, i.e.Â 4.0ÃvÃcosÎ¸=4.48
and
2.74â4.0ÃvÃsinÎ¸=0, i.e.Â 4.0ÃvÃsinÎ¸=2.74.
Dividing side by side we find tanÎ¸=2.744.48âÎ¸â31Â°. Having found the angle we can substitute in either of the two equations above to find v = 1.3 m sâ1.
The total kinetic energy before the collision was 12Ã2.0Ã6.02+0=36Â J and after the collision is 12Ã2.0Ã4.02+12Ã4.0Ã1.32=19Â J, so the collision was inelastic.
WORKED EXAMPLE 4.8
In the collision shown, the unknown quantities are the initial speed, w, of the red ball and the final speed, v, of the blue ball, Figure 4.23.

Figure 4.23: For worked example 4.8.

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Conservation of momentum gives
xÂ direction:Â 3.0Ãw+0îîîîîîîîîîîîîîtotalÂ before=3.0Ã5.0Ãcos60Â°+5.0ÃvÃcos20Â°îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotalÂ after
yÂ direction:0îîîîîîîîîîîîtotal beforeÂ =3.0Ã5.0Ãsin60Â°â5.0ÃvÃsin20Â°îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotal afterÂ 
These simplify to
3.0w = 7.5 + 4.698v
12.990 = 1.710v
From the second equation we find v = 7.5965 m sâ1. Substituting in the first, we find w = 14.3961 m sâ1.
We quote w = 14 m sâ1 and v = 7.6 m sâ1 as our answers.
WORKED EXAMPLE 4.9
The red ball has a glancing collision with a stationary blue ball of the same mass with the outcome shown (Figure 4.24).

Figure 4.24: For worked example 4.9.

Calculate v and Î¸.
Show that the collision is elastic.
Answer
Conservation of momentum gives
xÂ direction:Â mÃ8.0+0îîîîîîîîîîîîîîtotal beforeÂ =mÃ4.0Ãcos60Â°+mÃvÃcosÎ¸îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotal afterÂ 
yÂ direction:Â 0îîîîîîîîîîîîtotal beforeÂ =mÃ4.0Ãsin60Â°âmÃvÃsinÎ¸îîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîîtotal afterÂ 
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These simplify to
8.0=2.0+vcosÎ¸23â=vsinÎ¸i.e.,vcosÎ¸=6.0vsinÎ¸=23â
Dividing side by side gives tanÎ¸=23â6.0=3â3âÎ¸=30Â°. Then, v=6.0cos30Â°=6.03â2=123â=43ââ6.9Â mÂ sâ1.
The initial kinetic energy was EK=12mÃ8.02=32m. The final kinetic energy is EK=12mÃ4.02+12mÃ(43â)2=8m+24m=32m and so the collision is elastic.
TEST YOUR UNDERSTANDING
24In the collision in Figure 4.25, the two bodies have the same mass. What is u and what is v? Is the collision elastic? (Use exact values for sines and cosines.)

Figure 4.25: For question 24.

25Two bodies of equal mass collide (Figure 4.26). One body has speed w, and the other is at rest. After the collision the two bodies move such that the angle between them is 90Â°.

Figure 4.26: For question 25.

aShow that u = w cos Î¸ and v = w sin Î¸.
bHence show the collision is elastic.
26What are the initial and final speeds w and u of the red particle in Figure 4.27?

Figure 4.27: For question 26.

27Two bodies approach each other from opposite directions and collide, Figure 4.28.

Figure 4.28: For question 27.

What is the energy lost in this collision?
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28Two identical bodies moving with the same speed v collide as shown in Figure 4.29 and stick together. The two bodies move together with speed v2.

Figure 4.29: For question 28.

What is the angle Î¸?
NATURE OF SCIENCE
Physics and mathematics
General principles such as the conservation of momentum allow for simple and quick solutions to problems that may otherwise look complex. Consider, for example, a man of mass m who stands on a plank, also of mass m (Figure 4.30). There is no friction between the floor and the plank. The man starts walking on the plank until he gets to the other end, at which point he stops.
What happens to the plank?
The centre of mass must remain in the same place since there is no external force. So, the final position of the plank will be the one shown: the plank moves half its length to the left and stops.

Figure 4.30: The centre of mass remains in the same place.

