Reference no: EM13796070

1. In this week's experiment, you fired a ball into a pendulum and used the height that the pendulum reached to calculate the energy, and thus the velocity, of the ball when it was launched.  To do this, you had to use two relationships: one that described the relationship between kinetic energy and potential energy, and one that described the relationship between momentum before and after a collision.  The momentum equation you used was

mv₁ = (m+M)v₂

This equation, as you may have noticed, is a reduced version of the general momentum conservation equation,

mv_1,1 + Mv_1,2 = mv_2,1 + Mv_2,2

in which we assume two things: first, that the pendulum isn't moving before the collision; and second, that the collision between the ball and the pendulum is inelastic - that is, after the collision, the ball and the pendulum are stuck together and move together.  In an inelastic collision, very little energy is conserved.  If, however, the collision between the pendulum and the ball was totally elastic, all the energy would be conserved and the initial speed of the ball before the collision would be much closer to the initial speed of the pendulum after the collision.  As a result, the height the pendulum would reach would be greater than the height it reached in the experiment you ran.

a) Calculate how much higher the pendulum would swing if the collision was totally elastic instead of inelastic (that is, calculate the difference between the heights achieved through each collision). Assume that the ball's initial velocity v₁ as it is launched is 6.00 m/s, the mass of the ball m is .0666 kg, and the mass of the pendulum M is .275 kg.  For the inelastic collision, use the equation we used in class for the momentum relationship

mv₁ =  (m+M) v₂

For the elastic collision, we will assume that the ball transfers all its energy to the pendulum in the collision and then falls away so that the pendulum alone swings the rest of the way.  The equation in this case will be

mv₁ = Mv₂

In both cases the energy relationship after the collision will be

½mv₂= mgh

Where m represents whatever total mass you have after the collision and v is the velocity of the system after the collision.

b) Because the collision in the experiment was inelastic, a large percentage of the energy was lost as frictional heat, sound waves, etc.  Calculate the percentage of energy lost in the collision by comparing the energy the ball had before the collision to the energy that the ball and pendulum had after the collision. Assume that the ball's initial velocity v₁ is 6.00 m/s, the mass of the ball m is .0666 kg, and the mass of the pendulum M is .275 kg. Use the momentum relationship given above for the inelastic collision, and use the expression for kinetic energy to find your solution.

c) By now, you probably have noticed that momentum is always conserved in collisions, while energy is not always conserved in collisions.  In part a), we explored what would happen to the motion of the pendulum if the collision was elastic instead of inelastic.  In this part, I want you to find out what would happen if the collision remained inelastic but energy was somehow conserved (this is not actually physically possible).  First, find out how high the pendulum would swing if energy were conserved, such that the equation of energy would be

½mv₁² = ½ (m+M)v₂² = (m+ M)gh

Then, show how assuming energy conservation in this situation would violate the law of momentum conservation by showing that the momentum of the ball before the collision is not equal to the momentum of the ball and pendulum after the collision in that situation.

2. Another thing you did in this week's experiment was to predict where a ball launched from the spring would land, using the initial velocity you calculated from the first part of the experiment.  You can also use this process in the opposite direction - you can fire the launcher and measure where the ball lands to find out the initial velocity of the ball.

a) Say that you launch a ball with a mass m of .0666 kg horizontally from a launcher that is located at height y₀ 1.1 m above the floor. The distance you measure to where the ball lands x is located 1.8 m horizontally from the launch point.  (In this problem, y is used for vertical components and x is used for horizontal components.)  Use these relationships:

y = y₀ + v_0yt + ½ a_yt²

x = x₀ + v_0xt + ½a_xt²

And the knowledge you gained about their use from the lab to find out the initial velocity of the ball, v_0x.

(Note that in the equations I gave you, the values for several unknowns will be zero; part of your job in solving this problem is to know, or figure out, which of these unknowns are zero so that you can reduce the equations down to a more easily solvable form.  Also note that I have defined the coordinates in the y direction with up being positive and down being negative; make sure you take this into account when you are plugging the value for the acceleration due to gravity - a downward acceleration - into the y equation.)

b) Launching a ball horizontally is convenient because, as you found in the previous problem, it allows us to ignore several terms that would otherwise make solving the problem more complicated.  If you launch the ball at an angle, however, things get a little more complicated.  The initial speed of the ball will remain the same, but now part of the velocity will be in the y direction as well as the x direction. 

If you launch a ball with an initial velocity of 6.00 m/s at an angle of 45°, you can find the components of the velocity in both the vertical and the horizontal directions using the equations

v_0y = v₀ sinθ

v_0x = v₀ cosθ

Find these values and then plug them back into the kinematic equations given above to find out how far the ball will go (find x). 

Is this shorter, longer, or the same as the range you would get if you shot the ball horizontally (think about the experiment, which shot the ball horizontally at around 6.00 m/s)?

Why would this be the case?