Class Notes Physics I, 10/07/98

Momentum in Two Dimensions

We set the problem up as shown below.

• We will use v to stand for the unknown velocity of the 5 kg mass before collision.
• Since the velocity of the combined mass after collision is in the original direction of motion of the second mass, which is opposite to the (positive) original direction of the first mass, its velocity will be -2 m/s.

The known momenta are the 50 kg m/s of the first object before collision and the -20 kg m/second total momentum after collision.

• The -70 kg difference is accounted for as the momentum of the second object before collision.
• That is, since the total momentum before collision must equal the total momentum after collision, we see that the momentum of the second object before collision must be -70 kg m/s. 
• From this we can easily conclude that the second object has a velocity of -14 m/s before collision.

Proceeding a bit more formally, we set up the equation with the total momentum before collision set equal to the total momentum after collision.

• We easily solve this equation for the unknown velocity v of the first object before collision, obtaining v = -14 m/s.

We can solve this equation for v2B, the unknown velocity of the second object before collision.

• The numerator of the resulting expression is the difference between the total momentum after collision and the momentum of the first object before collision, which represents the momentum of the second object before collision.
• The denominator, which is the mass of the second object, shows that we divide the total momentum of the second object by its mass to obtain its velocity.

http://youtu.be/Or0FCzUvQ3Q

Momentum in Two Dimensions

• We will therefore model this situation in two dimensions, using x and y components of velocity and momentum.
• Since we can no longer indicate the direction of motion by + or -, as we could when the motion was all along a single axis, we will have to use vectors to express velocities and momenta.
• In this situation, where motion has components in the x and in the y directions, a velocity or momentum vector will have both x and y components.
• A velocity vector v (note that vectors will be indicated by bold-faced symbols, or alternatively by symbols with small arrows above them as below) will have components vx and vy in the x and y directions.
• A momentum vector p will similarly have x and y components px and py.

The situation in which a moving ball collides with a stationary ball is depicted below.

• The collision is such that the moving ball strikes the stationary ball to the right of center of the stationary ball, as it would be observed from the moving ball.
• After collision the moving ball's velocity will be deflected to the right and that of the stationary ball will be deflected to the left.
• The forward velocity of the first ball will be reduced as a result of the increased forward velocity of the second.

The geometric relationship among vx, vy and v is depicted by a rectangle formed by side of lengths vx and vy originating at a common point and pointing in the x and y directions, with the vector v representing the velocity running diagonally from the common point to the opposite diagonal of the rectangle.

• The rectangle is completed by dotted lines parallel to the x and y axes, running from the point of v to the points of vy and vx.
• These dotted lines are called projection lines.
• (Note that in the figure as shown below the shape of the rectangle has been somewhat distorted by a combination of camera angle and poor artwork on the part of the instructor; the angles between the projection lines and the component vectors vx and vy should be right angles).

From the geometric relationships and the definitions of the sine and cosine functions, we can see how the components vx and vy are related to the magnitude and angle of the vector v.

• A velocity vector v, representing motion in a direction which makes an angle `theta with respect to the positive x axis, will have x component vx = v cos(`theta) and y component vy = v sin(`theta), where the non-boldfaced v (or v without an arrow above it) stands for the magnitude of the velocity, or the speed, of the object.
• If the x and the y components of the velocity are known, then the magnitude v of the velocity is found by the Pythagorean Theorem to be v = `sqrt(vx^2 + vy^2) and the angle with respect to the x axis is `theta = tan^-1(vy / vx) or tan^-1(vy / vx) + 180 deg, depending on whether vx is positive or negative, respectively.

If we collide the two balls near the edge of a table of known height, so that their x and y velocities can be inferred from the x and y displacements of their landing locations with respect to the location of a straight drop from the point of collision (more specifically, from the location of the center of mass of the two balls at the instant of collision), we will obtain a picture like the one below.

• Here the x and y axes are both in the plane of the floor (this means that we are not regarding the vertical direction as the y direction; we might now regard the vertical direction as the z direction).

Knowing the z displacement from the edge of the table to the floor, we can easily infer the time required to fall through that displacement from rest (we assume that we have taken care that the vertical, or z velocities of the balls are 0 immediately after collision).

• Then by measuring the x and y displacements of the balls from the straight-drop position we can easily determine the x and y velocities of the balls immediately after collision.

If we know the masses of the balls, we can now easily find their momentum components p1x, p1y, p2x and p2y after collision.

• If the direction of the first ball before collision is taken as the x direction, we see that before collision that total momentum in the y direction was zero.
• It must follow that the total y momentum after collision must also be 0, so that p1y and p2y are equal and opposite.
• The total p1x + p2x of the momenta after collision must be equal to the momentum of the first ball before collision.

From the x and y components of the momenta after collision, we can find the magnitude and direction of the momentum vectors p1A and p2A after collision.

• These momenta can be expressed by the vectors p1A and p2A, as they might be depicted in the figure below.
• The total momentum after collision can be depicted by the sum of the two vectors, depicted by placing the vectors head-to-tail as shown below.

The total momentum after collision will be the vector originating at the 'tail' of the first vector and ending at the 'head' of the second.

• This total momentum should be equal to the total momentum pTotB before collision, which is completely in the x direction, as depicted by the vector pTotB in the figure below.

A more detailed sketch of the vector sum shows the components p1Ax, p1Ay, p2Ax and p2Ay of the two momenta after collision.

• Since the total momentum after collision should be zero, the y components should be equal and opposite.

The process of adding the x and y components is represented below, with the x and y component vectors all originating from the same point.

• This picture shows how the y components are equal and opposite and hence cancel (e.g., if instead of momentum components we imagine two equal and opposite force components acting on the same point, it is clear that neither force will win and that the net result will be 0).
• The x components of the momentum are in the same direction and hence reinforce one another, resulting in a total x component which is greater than either.
• The sum of these x components is depicted by the green vector below (this vector should have been drawn longer; its length should be equal to the sum of the lengths of p1xA and p2xA).