In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously.

One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin.

We start by assuming that [latex]\vec{\text{F}}_{\text{net}}=0,[/latex] so that momentum [latex]\vec{\text{p}}[/latex] is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 1.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 1. Because momentum is conserved, the components of momentum along the x– and y-axes (p_x and p_y) will also be conserved, but with the chosen coordinate system, p_y is initially zero and p_x is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.)

Along the x-axis, the equation for conservation of momentum is

Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is

But because particle 2 is initially at rest, this equation becomes

The components of the velocities along the x-axis have the form v cos θ. Because particle 1 initially moves along the x-axis, we find v_1x = v₁.

Conservation of momentum along the x-axis gives the following equation:

where θ₁ and θ₂ are as shown in Figure 1.

Along the y-axis, the equation for conservation of momentum is

or

But v_1y is zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v_2y is also zero. The equation for conservation of momentum along the y-axis becomes

The components of the velocities along the y-axis have the form v sin θ.

Thus, conservation of momentum along the y-axis gives the following equation:

The equations of conservation of momentum along the x-axis and y-axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.

Example 1: Determining the Final Velocity of an Unseen Object from the Scattering of Another Object

Suppose the following experiment is performed. A 0.250-kg object (m₁) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (m₂). The 0.250-kg object emerges from the room at an angle of 45.0° with its incoming direction, as shown in Figure 2. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′₂ and θ₂) of the 0.400-kg object after the collision.

 

 

Strategy

Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 2 is one in which m₂ is originally at rest and the initial velocity is parallel to the x-axis, so that conservation of momentum along the x– and y-axes is applicable.

Everything is known in these equations except v′₂ and θ₂, which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the x– and y– directions.

Solution

Solving m₁v₁ = m₁v′₁ cos θ₁ + m₂v′₂ cos θ₂ for v′₂ cos θ₂ and 0 = m₁v′₁ sin θ₁+m₂v′₂ sin θ₂ for v′₂ sin θ₂ and taking the ratio yields an equation (in which θ₂ is the only unknown quantity. Applying the identity [latex](\tan\theta=\frac{\sin\theta}{\cos\theta}),[/latex] we obtain:

[latex]\tan\theta_2=\dfrac{v^{\prime}_1\sin\theta_1}{v^{\prime}_1\cos\theta_1-v_1}.[/latex]

Entering known values into the previous equation gives

[latex]\tan\theta_2=\dfrac{(1.50~\text{m/s})(0.7071)}{(1.50~\text{m/s})(0.7071)-2.00~\text{m/s}}=-1.129.[/latex]

Thus,

[latex]\theta_2=\tan^{-1}-1.129=311.5^0\approx312^0.[/latex]

Angles are defined as positive in the counter clockwise direction, so this angle indicates that m₂ is scattered to the right in Figure 2, as expected (this angle is in the fourth quadrant). Either equation for the x– or y-axis can now be used to solve for v′₂, but the latter equation is easiest because it has fewer terms.

[latex]v^{\prime}_2=-\dfrac{m_1}{m_2}v^{\prime}_1\dfrac{\sin\theta_1}{\sin\theta_2}[/latex]

Entering known values into this equation gives

[latex]v^{\prime}_2=-(\dfrac{0.250~\text{kg}}{0.400~\text{kg}}) 1.50~\text{m/s}-(\dfrac{0.7071}{-0.7485}).[/latex]

Thus,

[latex]v^{\prime}_2=0.886~\text{m/s}.[/latex]

Discussion

It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further.