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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 θ 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 θ.

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.

∑p y⏟before =∑p y⏟after

and similarly for the y direction,

∑p y⏟before =∑p y⏟after

Applying these equations to our collision we find

x direction: :2.0×6.0+0⏟total before =2.0×4.0×cos 20°+4.0×v×cos θ⏟total after

y direction: 0⏟total before =2.0×4.0×sin 20°−4.0×v×sin θ⏟total after

These imply

7.52+4.0×v×cos θ=12, i.e. 4.0×v×cos θ=4.48

and

2.74−4.0×v×sin θ=0, i.e. 4.0×v×sin θ=2.74.

Dividing side by side we find tan θ=2.74 4.48⇒θ≈31°. Having found the angle we can substitute in either of the two equations above to find v = 1.3 m s–1.

The total kinetic energy before the collision was 1 2×2.0×6.0 2+0=36 J and after the collision is 1 2×2.0×4.0 2+1 2×4.0×1.3 2=19 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

x direction: 3.0×w+0⏟total before=3.0×5.0×cos 60°+5.0×v×cos 20°⏟total after

y direction:0⏟total before =3.0×5.0×sin 60°−5.0×v×sin 20°⏟total after

These simplify to

3.0w = 7.5 + 4.698v

12.990 = 1.710v

From the second equation we find v = 7.5965 m s–1. Substituting in the first, we find w = 14.3961 m s–1.

We quote w = 14 m s–1 and v = 7.6 m s–1 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 θ.

Show that the collision is elastic.

Answer

Conservation of momentum gives

x direction: m×8.0+0⏟total before =m×4.0×cos 60°+m×v×cos θ⏟total after

y direction: 0⏟total before =m×4.0×sin 60°−m×v×sin θ⏟total after

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These simplify to

8.0=2.0+v cos θ v cos θ=6.0 i.e.,2 3=v sin θ v sin θ=2 3

Dividing side by side gives tan θ=2 3 6.0=3 3⇒θ=30°. Then, v=6.0 cos 30°=6.0 3 2=12 3=4 3≈6.9 m s−1.

The initial kinetic energy was E K=1 2 m×8.0 2=32 m. The final kinetic energy is E K=1 2 m×4.0 2+1 2 m×(4 3)2=8 m+24 m=32 m 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 90°.

Figure 4.26: For question 25.

aShow that u = w cos θ and v = w sin θ.

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 v 2.

Figure 4.29: For question 28.

What is the angle θ?

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.