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Kurs: Fizyka > Rozdział 6

Lekcja 1: Pęd i popęd

Czym są zderzenia w dwóch wymiarach?

Learn how to handle collisions in 2 dimensions...and get better at playing billiards.

How can we solve 2-dimensional collision problems?

In other articles, we have looked at how momentum is conserved in collisions. We have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy. We have applied these principles to simple problems, often in which the motion is constrained in one dimension.

If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).

For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).

In other words, the total momentum in the x direction will be the same before and after the collision.

Σ p_{x i} = Σ p_{x f}

Also, the total momentum in the y direction will be the same before and after the collision.

Σ p_{y i} = Σ p_{y f}

In solving 2 dimensional collision problems, a good approach usually follows a general procedure:

Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.
Write down all the values you know and decide exactly what you need to find out to solve the problem.
Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.
Identify all the forces acting on each of the bodies in the system. Make sure that all impulse is accounted for, or that you understand where external impulses can be neglected. Remember that conservation of momentum only applies in cases where there is no external impulse. However, conservation of momentum can be applied separately to horizontal and vertical components. Sometimes it is possible to neglect an external impulse if it is not in the direction of interest.
Write down equations which equate the momentum of the system before and after the collision. Separate equations can be written down for momentum in the x and y directions.
Solve the resulting equations to determine an expression for the variable(s) you need.
Substitute in the numbers you know to find the final value. Should this require adding vectors, it is often useful to do this graphically. A vector diagram can be drawn and the method of adding vectors head-to-tail used. Trigonometry can then be used to find the magnitude and direction of all the vectors you need to know.

Billiard ball problem

Figure 1 describes the geometry of a collision of a white and a yellow billiard ball. The yellow ball is initially at rest. The white ball is played in the positive-x direction such that it collides with the yellow ball. The collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis.

The mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg. A sound recording reveals that the collision happens 0.25 s after the player has struck the white ball. The yellow ball falls into the pocket 0.35 s after the collision.

Exercise 1a: What is the velocity of the white ball after the collision?

In this problem we need to know the velocity (magnitude and direction) of the white ball after the collision. We will use the subscripts w and y for the white and yellow balls, and i and f for initial and final. An x and y axis has been labeled in the figure, so we will use these as the axes in our analysis.

The figure shows the relevant distances of travel of each ball. We know the time between each event so we can find the velocities. Assembling all the information we know:

\begin{aligned} m_{w} & = 0.18 kg \\ m_{y} & = 0.15 kg \\ v_{wi} & = \frac{1 m}{0.25 s} ∠ 180^{\circ} \\ v_{yi} & = 0 \\ v_{yf} & = \frac{1 m}{0.35 s} ∠ 62^{\circ} \end{aligned}

There is no indication that energy is conserved in this problem. (It is an inelastic collision. All the impulse in the collision comes from the moving white ball.)

Collisions between rigid billiard balls happen fast, so we can apply conservation of momentum:

\begin{aligned} m_{w} \cdot v_{wi} & = m_{w} \cdot v_{wf} + m_{y} \cdot v_{yf} \\ v_{wf} & = \frac{1}{m_{w}} (0.72 ∠ 180^{\circ} kg \cdot m / s - 0.428 ∠ 62^{\circ} kg \cdot m / s) \end{aligned}

The vector arithmetic which is needed to find

v_{wf}

can be done graphically. Figure 2 shows a vector diagram with both momentum vectors

p_{wi}

and

- p_{yf}

added head-to-toe. The blue momentum vector connecting the start of the chain to the end represents the sum,

v_{wf}

in this case.

Note that subtracting the vector

p_{yf}

is the same as adding

- 1 \cdot p_{yf}

. Multiplying by -1 is achieved by simply flipping the direction of the vector. These articles on vectors contain many examples of similar vector manipulation.

The unknown values are solved using trigonometry and shown in blue. The resultant vector shown in blue is the final momentum of the white ball. Dividing by the mass of the white ball gives its velocity.

\frac{0.396 kg \cdot m / s}{0.18 kg} ∠ - {30.33}^{\circ} = 2.2 m / s ∠ - {30.33}^{\circ}

Note that it is also possible (and generally quicker) to do the vector arithmetic using a calculator which can work with complex numbers. However, it is still advisable to sketch a vector diagram so your result can be visually checked.

Exercise 1b: Is the white ball likely to fall into any of the pockets? If so, how could that be avoided?

The figure conveniently shows the angle of the line intersecting the white ball's center of mass at the collision and the upper right pocket. It is shown to be

31^{\circ}

from the x-axis. Since this is very close to our calculated angle, it is likely the ball will fall into this pocket.

As long as the direction the white ball is initially played remains the same, the contact point with the yellow ball will remain the same and it is guaranteed to fall into the top right pocket. If we either increase or decrease the magnitude of the initial velocity then the white ball will hit the upper or right-most edge of the table respectively.

Exercise 1c: How much energy is lost to the environment in this collision?

Bouncing baseball

Consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine. The machine is set to deliver the 0.145 kg ball at 10 m/s. The ball hits the board, making an angle of 45° from the surface of the board. The board is slightly flexible and the collision is inelastic. The ball bounces back at an angle of 40° from the surface of the board as shown in figure 3.

Hint: When a ball bounces off a surface, the impulse responsible for the bounce is always directed normal to the surface.

Exercise 2a: What is the velocity of the ball after the collision?

The only bodies in this system are the ball and the back-board. We will use the subscript b for the ball and subscripts i and f for initial and final. For our analysis we will use the axes shown in figure 3.

In this problem we know the initial momentum of the ball

p_{bi} = m_{b} v_{bi} = 1.45 kg \cdot m / s

at a

45 °

angle. We know the direction of the final momentum vector

p_{bf}

, but not the magnitude of this vector. Indeed, this magnitude is what we need to find to solve the problem.

In this problem the back-board is considered to be stationary and its elastic properties are unknown. The impulse it delivers to the ball during the collision is an unknown external impulse. However, we do know that this impulse is directed normal to the surface (-y direction).

There is no external impulse in the x direction, so momentum is conserved in this dimension. If the change in momentum due to the back-board is

Δ p_{board}

, then by conservation of momentum:

p_{bf} = p_{bi} + Δ p_{board}

Problems like this are often easiest to solve graphically. Figure 4 shows a momentum vector diagram with the vectors on the right hand side of the equation added head-to-toe and shown in blue. The resulting sum vector

p_{bf}

is shown in green. Where the magnitude of a vector is initially unknown, a dashed line has been extended. It is possible to find the magnitude of the unknown vectors by finding where the lines intersect using trigonometry.

The ball's final momentum vector is determined by first calculating the length of adjacent side of the upper triangle, then the hypotenuse of the lower triangle as shown. The velocity is then

\frac{1.338 kg \cdot m / s}{0.145 kg} = 9.227 m / s

Exercise 2b: If the ball is in contact with the wall for 0.5 ms, what is the magnitude of the force on the ball due to the wall?

Exercise 2c: How much work was done on the wall by the ball?

Although the backboard does not move, it is slightly flexible. When the ball hits the wall and rebounds, the wall flexes, so there is work done on the wall by the ball. Small pieces of the wall near the ball indeed move, so there is a force applied by the ball over a distance, which means that work is done on the wall by the ball. This work

W

done on the back board is the change in kinetic energy of the ball. Afterward, that energy absorbed by the backboard would be lost to the surrounding environment as sound or heat.

We can subtract the final kinetic energy from the initial to find the total work

W

the ball must have done to the wall.

\begin{aligned} W & = \frac{1}{2} m_{b} v_{bi}^{2} - \frac{1}{2} m_{b} v_{bf}^{2} \\ = \frac{1}{2} 0.145 kg (10 m / s)^{2} - \frac{1}{2} 0.145 kg (9.227 m / s)^{2} \\ = 1.077 J \end{aligned}

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