Apparatus: Steel ball, glass ball, a telephone with slow motion camera, and a level glass table.
Purpose: Look at a two-dimensional collision and determine if momentum and energy are conserved.
Model: P0=Pf and P=mv, 1/2m1v0^2=1/2m1v1^2+1/2m2v2^2
P is the momentum, P0 is the initial momentum, Pf is the final momentum, m is the mass of the object, and v is the velcity of the object.
Process: We weigh the mass of the glass ball and the steel ball, Mg=0.0196kg and Ms=0.069kg. First, we level the the glass the table and place a telephone on the top of the table. We gently set the stationary glass ball on that level table, and we use another glass ball to aim the stationary glass ball so that it could hit the side of the stationary ball. We start the slow motion camera to record, and we push the ball to hit the stationary ball. Then, we use the glass ball as the stationary ball, but we use a steel ball to be the rolling ball.
After recording these two vedio, we upload these vedio from the telephone to the computer. We use the software Quick to edit the vedio. Then, we open the vedio by using the Logger Pro. We change the frame rate to 240fps. We add the point on the vedio to mark the motion of the balls.
Then, we add the new column about the momentum of rolling ball in x-axis and y-axis and the momentum of stationary ball in x-axis and y-axis by using P=mv.
As the graph show, in two different experiment, the total momentum of two balls is constant in x-axis and y-axis. The total momentum does not change during the collision. The power of the entire system should not be change during the elastic collision. However, the power in the entire system is not constant, and it decrease a little bit when the balls hit each other; therefore, we can know that this collision is not an elastic collision. There are some power changing to other energy form.
We use the Logger Pro to find out the center of mass of entire system. Then, we figure out that the motion of center of mass is a straight line, and the momentum of the center of mass is constant.
Apparatus: a spring-loaded "gun" and a nylon block with rubber in it
Purpose: Determine the firing speed of a ball from a spring-loaded-gun.
Model: vombullet=vfmtotal, 1/2gt^2=h, x=vft, and 1/2mtotalvf^2=(1-cosθ)Lmtotalg
vo is the velocity of the bullet flying out the barrel, mbullet is the mass of the bullet, vf is the velocity of the block-bullet, t is the time that the bullet hit the ground, h is the heigh from the barrel to the ground, X is the distance from the barrel to the hit point on the ground, θ is the angle the pendulum hit the highest point, and L is long of the string.
Purpose: We measure and record the mass of the bullet mbullet and block mblock, mbullet=0.0075kg and mblock=0.0795kg. We determine the length of the string of the pendulum L, L=0.205m. We level the base of the apparatus to make the barrel can fire horizontally and make sure the block is level.
After making sure the bullet can hit the block and strick into the block, we pull back and lock the spring into the first notch, put the angle indicator to zero degress, and place the ball into the barrel.
We push the locker and fire the bullet into the block. The block push the angle indicator back. We read and record the maximum angle θ, θ=33°.
Then, we determine the heigh from the barrel to the ground h, h=0.99m. Without the block, we fire the bullet. it fly and hit the ground. We use the carbon record paper to mark the hit point. Then, we determine the horizontal distance between the barrel and the hit point x, x=2.36m
Then,
vo1 is the initial velocity of the bullet which we gets by shooting the bullet to the ground
1/2gt^2=h
t=0.449s
vo1t=x
vo1=5.25m/s
vo2 is the initial velocity of the bullet which we gets by shooting the bullet into the block
vo2mbullet=vfmtotal
1/2mtotalvf^2=(1-cosθ)Lmtotalg
vf=0.492m/s
vo2=5.71m/s
We get vo1≈vo2, considet that the uncertainty for every data that we determine. We assume they are the same.
Apparatus: An air-track glider, an air track, a strong magnet, a motion detector, a telephone with an angle detector.
Purpose: Verify that conservation of energy applies to this system.
Model: Fmag=sinθmg, Fmag=Ax^B, and W=-∫Fmagdx=(A/(B+1))x^(B+1)
Fmag is the force existing on the magnet, θ is the angle of the track, m is the mass of the air-track glider, A and B is the constant of the magnet, x is the separation between the glider and the magnet, and W is the potential energy on the magnet.
Process: The air-track glider with a strong magnet on one end approaches a fixed magnet of the same polarity on the other end.
We raise one end of the air track which has the glider and record the angle θ. The glider slide will begin to slide down and stop at a position because of the force giving by magnet. Then, we determine the separation between the glider and the magnet x.
After six times changing the angle θ, we get:
θ 3.7° 7.6° 10.3° 13.3° 15.1° 17.4°
x 18.2 13.6 11.3 10.4 9.1 8..7
By using the model Fmag=sinθmg, we get:
F 0.214 0.442 0.594 0.764 0.865 0.993
We plot x as x-axis and F as y-axis into the LabPro, and we use the Curve Fit--Ax^B to get:
According to the graph, we get:
Fmag=0.00003766x^-2.168
Then, we place the the air track horizontally, and the motion detector is placed on the magnet side. We record the distant that the motion detector read L and the separation on that position l. We can use
x=xread-(L-l) to get the actul separation.
We creat a new columm about W by using W=-∫Fmagdx=(A/(B+1))x^(B+1)=0.000032243x^-1.168,
KE by using KE=(mv^2)/2, and Esum by using Esum=KE+W.
We push the glider toword the motion detector, and we start use the computer to collect the data. Then we get:
When we push the glider, we give the glider some kinetic energy. Without friction, the kinetic energy should be constant until getting close enough to the magnet. When the glider get close to the magnet, the kinetic energy decrease, and the potential energy increase. The kinetic energy transfer to potential energy. Then, after the kinetic energy reach the zero, the kinetic energy begin increasing, and the potential energy begin decreasing. The potential energy transfer to kinetic energy.
The sum of the eneegy should be constant and equal to the energy that we get the glider by pushing.
Apparatus: Two clamps, a metal rod, a motion detector, a spring, and some masses.
Purpose: Find out the relationship about the energy in a vertically-oscillating mass-spring system.
Model: KEhanging=(mhanging*v^2)/2, KEspring=(mspring*v^2)/3, GPEhanging=mhanginggh, GPEspring=mspringgh/2, Elastic PE=x^2k/2, x=H-h. KEtotal=KEhanging+KEspring, and GPEtotal=GPEhanging+GPEspring.
KEhanging is the kinetic erengy of hanging mass, KEspring is the kinetic energy of spring, GPEhanging is the gravitational potential energy of hanging mass, GPEspring is the gravitational potential energy of spring, Elastic PE is elastic potential energy of the spring, mspring is the mass of the spring, mhanging is the mass of the hanging mass, v is the velocity of the hanging mass, h is the heigh of the mass, x is the length of the spring, and H is the heigh of the top of the spring.
Process: First, we determine the mass of spring and the mass of the masses. We use the clamps and the meter rod to hang a spring vertically. In the bottem of the spring, we hang some masses. We place a motion detector directly below the mass and the spring. Then, we measure the heigh of the top of the spring H.
The apparatus set up as:
Then, we pull the masses down, stretch the spring, and let it go. click the strat buttom to begin collecting data such velocity v. the heigh of the masses h.
We add the calculate column of KEhanging, KEspring, GPEhanging, GPEspring, Elastic PE, KEtotal, GPEtotal by using the equation KEhanging=(mhanging*v^2)/2, KEspring=(mspring*v^2)/3, GPEhanging=mhanginggh, GPEspring=mspringgh/2, Elastic PE=x^2k/2, x=H-h. KEtotal=KEhanging+KEspring, and GPEtotal=GPEhanging+GPEspring.
We get the graph and data:
We plot KEtoyal, GPEtotal, Elastic PEtotal and Esum vs. position on one graph. We get:
The sum of the energy is the sum of the kinetic erengy of hanging mass, the kinetic energy of spring, the gravitational potential energy of hanging mass, the gravitational potential energy of spring, and elastic potential energy of the spring. The sum of the energy should be a constant number. This constant number should due to the length that the spring be stretched. It is the initial energy that the entire system has.
However, in our experiment, the sum of the energy is not constant. This error may cause by the spring constant of the spring that we measure in last experiment or the heigh of the top of the spring that we measure.
Apparatus: A ramp, a cart, a motion detector, force sensor, a spring, and some masses.
Purpose: Find the relationship between work done by force and the kinetic energy, and find the spring constant of the spring.
Model: ΔW=ΔKE, W=Fx, KE=(mv^2)/2, and F=kx.
W is the work done on cart by force, KE is the kinetic energy of the cart, m is the mass of the cart, the velocity of the cart, k is the spring constant.
Process: First, we set up the apparatus as showed on picture. We placed the ramp on the horizontal table, and we put the motion detector on one side of the ramp and the force sensor on the other side of the ramp. We put the cart on the ramp, and then we collect the cart and the force sensor by using a spring. The spring needed to be kept on horizontal.
We measured the mass of the cart m. m=0.71kg. We opened the LabPro, calculated the force sensor, and zero the force sensor and the motion sensor. Then, we created the new calculated column for the kinetic energy, KE, by using KE=(mv^2)/2 and the work, W, by using W=Fx.
We pull the cart toward the motion detector, started collecting the datas, and let the cart go.
Then, we got the graph:
We selected the part of the graph that we needed and used the integration routine to the Force vs. Position.
We chose x=0.243m W=0.3297J KE=0.318J
x=0.190m W=0.4614J KE=0.427J
x=0.130m W=0.5662J KE=0.506J
The W is very similar to KE, but they still have some difference, This error may cause by the work done by friction.
By reading the slope of Force vs. Position, k=11.64N/m.
Apparatus: An electric motor mounted on a surveying tripod, a long shaft going vertically up from the shaft, a horizontal rod mounted on the vertical rod, a long string tied to the end of the horizontal rod, a rubber stopper at the end of the string, and a ring stand with a horizontal piece of paper or tape sticking out.
Purpose: Find the relationship between θ and ω, and compare the theory value of ω and the experiment value of ω.
Model: ω=2π/t, mrω^2=F, cosθ=(H-h)/L, F=tanθmg, r=tanθL+R
ω is the angular speed of the stopper, θ is the angle between the string and vertical, t is the period of the rotation, m is the mass of the stopper, H is the heigh of the rod, h is the heigh of the stopper, L is the length of the srting, F is the centripetal force, r is the radius of the rotation, and R is the length of the rod.
Process: Our professor set up the apparatus as showed on the photo below.
We determined the heigh of the rod and the length of the string before it started. Then, professor turn on the motor. The stopper started rotating. By changing the power of the motor, we collected the different heigh of the stopper h and the time of 10 rotations the stopper 10t.
We used
ωtheory=2π/t
to find out the theory value of angular speed, and we used
ωexperiment=(tanθg/(sinθL+R))^(1/2), θ=cos^-1((H-h)/L)
to find out the experiment value of angular speed.
Compare the ωtheory and ωexperiment and find out the off between them.
There are some error during the expriement such as the rod (meter strick) does not keep horizontal when it is rotating. The rod's changing makes the heigh H different than we think, so the angle θ is different, the radius is different, and the centripetal force become different.
And when we determine the heigh of the stopper, it makes the stopper slower because it hits the paper. They make ωexperiment. The slower speed make ωtheory larger.
Apparatus: A big disk, some masses, a motor, and a force sensor,
Purpose: To determine the relationship between centripetal force and mass* angular velocity^2; centripetal force and radius* angular velocity^2; centripetal force and angular velocity^2.
Model: F= mrw^2
F is the centripetal force, m is the mass of the mass, and w is the angular velocity of the mass or disk.
Process: Professor set up the apparatus. He puts the force sensor on the center of the disk and ties a mass on the force sensor by using a long string. On the bottom of the disk, a motor is connected on the center of the disk to make the disk rotate.
Then, by changing the mass, the length of rope, or the angular speed, we collect some data:
the mass m, the radius r, the time of the disk for ten round t, and the force f
r m t w f
0.205 0.2 15.7 4.002 0.6844
0.295 0.2 15.39 4.083 1.13
0.41 0.2 14.37 4.372 1.67
0.58 0.2 14.33 4.385 2.35
0.58 0.2 12.38 5.075 3.13
0.58 0.2 12.47 4.665 3.83
0.58 0.2 10.16 6.184 4.753
0.58 0.1 10.58 5.939 2.16
0.58 0.05 10.74 5.850 1.03
Keeping the r be constant, we plug f vs. mw^2 into the Logpro to get the graph. According to the model f = mrw^2, in the graph f vs. mw^2, the slope of the graph is the radius r. r= 0.6411m. The actual raduis r=0.56m. These datas are different, and the determent's error and the friction between the disk and the mass may cause this error.
Keeping the m be constant, we plug f vs. rw^2 into the Logpro to get the graph. According to the model f = mrw^2, in the graph f vs. rw^2, the slope of the graph is the mass m. m= 0.2052kg. The actual rmass m=0.2kg. These two datas are very close that mean we are doing well at this group of data.
Keeping the mr be constant, we plug f vs. w^2 into the Logpro to get the graph. According to the model f = mrw^2, in the graph f vs. w^2, the slope of the graph is mass*radius mr. mr=0.129kg*m. The actual mass*radius mr=0.116kg*m. These two datas are close, but they are not close enough. The determent's error and the friction between the disk and the mass may cause this error.
The mass m has uncertainty 0.01 kg, the radius r has uncertainty 0.001, the time of the disk for ten round t has uncertainty 0.01, and the force f has uncertainty 0.001. They cause about 1% off for the calculation.
