Energy, Vectors and Projectile Motion

Class Notes Physics I, 10/28/98

Energy, Vectors and Projectile Motion

The quiz problem asked for the magnitude and angle of the velocity of a projectile as it strikes a ground, provided that the projectile starts from a height of 20 meters with a velocity of 15 meters per second in the horizontal direction.

We observe that in order to find the magnitude and angle of the final velocity, we need only find the final x and y velocities. Once these are known we can easily find the magnitude and angle of the final velocity.

We can easily find the two components of the final velocity.

The x component is of course the easiest; since we know that the initial x velocity is 15 meters/second, and since we assume for projectile motion that the horizontal velocity is unchanged, the horizontal component of the final velocity will be 15 meters/second.

The final y velocity requires a little more work, but can still be found in a straightforward manner.
The initial y velocity is zero (since the initial velocity is horizontal), the y acceleration is -9.8 meters/second^2, and the y displacement is -20 meters.
Using these quantities we easily determine that the final y velocity is -19.8 meters/second. These calculations and observations are summarized below.

From vfx = 15 m/s and vfy = -19.8 m/s, we find that the final velocity is 25 meters per second at an angle of -53 degrees.

If we want a positive angle between zero and 360 degrees is desired, we can express -53 degrees as 360 - 53 degrees = 307 degrees.

http://youtu.be/C4RFg2S3Qus

We now look at the energies involved in this situation.

Since the mass of the projectile is not known, we use M to stand for its mass (we use the capital letter to distinguish mass from meters in this situation).
We easily see that the initial kinetic energy is .5 M (15 m/s) ^ 2 = (112.5 m^2 / s^2) * M and, from the 25 meter/second magnitude of the final velocity, we see that the final kinetic energy is approximately (312 m^2/s^) * M.
Thus we have a change of approximately (200 m^2 / s^2) * M in the kinetic energy, as observed from the instant the ball leaves the lead to the instant it strikes the ground.

The potential energy change between the two corresponding altitudes is equal to - M g (20 m) = -M (196 m^s / s^2), which is approximately -(200 m^2 / s^2) * M.

We thus see that the increase in kinetic energy is equal to the decrease in potential energy.
We see this as a result of energy conservation; when the potential energy is 'lost', the 'lost' energy is 'found' in the kinetic energy increase.
The total energy of the system thereby remains constant, as the Law of Energy Conservation dictates.

There are two other ways to express this.

Since `dKE = (200 m^2/s^2) * M and `dPE = -(200 m^2 / s^2) * M, we see that `dPE + `dKE = 0.
This is, of course, equivalent deciding that PE + KE = constant.

http://youtu.be/llI6Ueg_NIA

We now consider a situation in which the object has initial velocity 40 m/s at an angle of 30 degrees above horizontal, with all other quantities as in the previous example.

We could analyze this situation by finding our initial vertical and horizontal velocities, which are v0y = 40 m/s * sin(30 deg) = 20 m/s and v0x = 40 m/s * cos(30 deg) = 35 m/s (approximately).
We could then analyze the vertical horizontal motion in the usual manner.
However, in this example we will use energy to determine the magnitude in angle of the final velocity.

We easily see that the initial kinetic energy is (800 m^2 / s^2) * M, and that the change in potential energy is still approximately (-200 m^2 / s^2) * M.

We thus conclude that the kinetic energy will change by + (200 m^2 / s^2) * M, giving us a final kinetic energy of (1,000 m^2 / s^2) * M.
If we set this equal to .5 M vf^2, we easily obtain vf = 44 m/s.

In the figure below we obtain this result somewhat more formally, using `dKE + `dPE = 0.

If the final velocity is vf, then `dKE = KEf - KE0 = .5 m vf^2 - (800 m^2 / s^2) * M, and statement of `dKE + `dPE = 0 is as shown below.

In the figure below we use -(196 m^2 / s^2) * M instead of -(200 m^2 / s^2) * M for the change in potential energy.

This results in an equation which is easily solved for vf: we first divide both sides by M, then proceed in a straightforward manner.

At this point we know that vfx = 35 m/s (this is equal to the initial x velocity, since x velocity is on changing), and that v = 44 m/s.

Since by the Pythagorean Theorem v^2 = vfx^2 + vfy^2, we can easily find vy, as shown in the partially erroneous figure below.

Having found vfy, we can easily find the angle of the final velocity.

Note errors: The errors in this figure are as follows: vy should be labeled vfy, and this velocity should be negative. We should have written vfy = +- `sqrt(v^2 - vx^2), and noted that we choose the negative value since the final y velocity is clearly negative. This negative velocity would have given us and angle of tan^-1(-28 m/s / (35 m/s)) = -34 deg.

http://youtu.be/hkFLuEDUYL4

We could also have used energy to determine the maximum altitude to which the projectile rises.

At its maximum altitude, the projectile's velocity is in the horizontal direction.
The velocity at that point is therefore just the constant 35 m/s horizontal velocity, and the kinetic energy is easily found to be approximately (613 m^2/s^2) * M.
Since the initial kinetic energy was (800 m^2/s^2) * M, this represents a kinetic energy change of approximately -(187 m^2/s^2) * M, as indicated in the figure below.
This KE change would be associated with an equal and opposite PE increase.
Since the change in potential energy is `dPE = Mg `dy, where `dy is the altitude change, we see that `dy = `dPE / (Mg) = (187 m^2/s^2) * M / (M g) = 19 m (approximately).

http://youtu.be/hYEn6mD5V00

Text-related Problems

A question was raised about a (Level III !!) text problem in which we were to derive an expression for the angle of the velocity at clock time t, relative to the horizontal direction, of a projectile whose initial velocity at t = 0 was v0 in the horizontal direction.

In light of the preceding examples this is an easy problem. We need only find expressions for the x and y components of the velocity at time t.
The horizontal velocity is clearly vx = v0, since horizontal velocity is unchanging.
Since the magnitude of the acceleration in the vertical direction is just the acceleration of gravity, g, the change in vertical velocity will be - g `dt = - gt.
Since the original vertical velocity was 0, the vertical velocity at time t is just vy = - gt.

Now since vx = v0 and vy = - g t, the angle at which the projectile is moving will be `theta = tan^-1 ( vy / vx) = tan^-1(-gt / v0).

Though it was not requested, we note that the magnitude of the velocity at time t will be v = `sqrt(v0^2 + (g t) ^ 2). This implies a kinetic energy of .5 * (v0^2 + (g t)^2) * M.

This kinetic energy could also have been found from the conservation of energy:
The initial kinetic energy was .5 v0^2 * M; since the vertical displacement is - .5 g t^2, the change in potential energy will be M g `dy = M g (-.5 g t^2) = -.5 M (g t)^ 2, implying a kinetic energy increase of .5 M (g t) ^ 2.
The kinetic energy at time t will therefore be the sum of the initial kinetic energy and the kinetic energy change, or .5 v0^2 * M + .5 M (g t)^2 = .5 * (v0^2 + (g t)^2) * M.

http://youtu.be/XXcoZfMpNew

The figure below depicts a 2500 kg tractor pulling a 1500 kg sled full of rocks, behind which the very large owner of the rocks is pulling by a rope while dragging his feet.

The coefficient of kinetic friction between the sled and the road surface is .3, while that between the sturdy shoes of the owner of the rocks is .8.

The tractor is accelerating at .7 m/s^2.

We wish to find the tension in each of the chains, and also the minimum possible coefficient of friction between the wheels of the tractor and the surface over which is rolling.

We begin by regarding the owner of the rocks and the sled as a system. The total mass of the system is 1750 kg.

Assuming that the road surface is horizontal, the vertical forces acting on the system will be

the 14,700 Newton weight of the sled, directed downward, and the equal and opposite normal force exerted by the road

the 2450 Newton weight of the individual, also directed downward, and the equal and opposite normal force exerted on him by the road.

The normal forces will give rise to frictional forces f1 = (14,700 N) * .3 = 4400 N (approximately) and f2 = 2450 N * .8 = 2000 N (approximately), both directed in the negative x direction (we choose the x direction as the direction to the right, which is the direction which the acceleration occurs).

The only other force acting on the system is the tension in the rope between the tractor and the sled. This situation is depicted above.

We note that the net force is entirely in the x direction and is equal to T - f1 - f2 = T - 4400 N - 2000 N = T - 6400 N.
We note also that the net force results in an acceleration of .7 m/s^2. Since the total mass of the system is 1750 kg, this net force must be net force = .7 m/s^2 * 1750 kg = 1200 N (approximately).
We thus have net force = T - 6400 N = 1200 N, which we solve for T to obtain T = 7600 N.

The coefficient of friction required between the tractor and the road is therefore the ratio 7600 N / 24,500 N = .3 (approx.) of the force it must exert to the normal force between the tractor and the road surface.

The tension in the second rope can be found by a process identical to the one used above, where in this case the system consists only of the 250 kg individual.

Alternatively, we can reason the result of more intuitively as follows:

The tension must be sufficient to overcome the 2000 N frictional drag and in addition to accelerate the individual at .7 m/s^2.

This acceleration implies a net force of 250 ks * .7 m/s^2 = 175 N, so a tension of 2175 N is required to supply the net force and overcome the drag.

http://youtu.be/p1VeoZqcYMU