Inelastic Collisions

Introduction


In this lab we analyzed the motion of two low friction carts during an inelastic collision and verified that the law of conservation of linear momentum was obeyed. We used the carts and track as shown in the figure.  If we regard the system of the two carts as an isolated system, the momentum of this system will be conserved.  If the two carts have a perfectly inelastic collision, that is, stick together after the collision, the law of conservation of momentum says that the initial momentum is equal to the final momentum.

Calculations

Pi = Pf
m1v1 + m2v2  =  (m1 + m2)V

Pi = m1v1 + m2v2 = (0.5060 * 0.4825)  + (0.4995 * 0.4825) = 0.2441 kg * m/s

Pf = ( m1 + m2)V = (0.5060 + 0.4995) * 0.1911 = 0.1922 kg * m/s

 v1, v2 = velocities before the collision
V = velocity of the combined mass after the collision.

Data

With the second cart (m2) at rest give the first cart (m1) a moderate push away from the motion
detector and towards m2.

Place an extra 500 g on the second cart and repeat

Remove the 500 g from the second cart and place it on the first cart and repeat

Trial	Percent Difference (%)
1	21.3
2	15.61
3	10.06
4	3.83
5	9.31
6	8.27
7	3.55
8	4.12
9	3.87
Average	8.88%

Theoretical calculations for ΔK/K in a perfectly inelastic collision of three

situations:

1. a mass, m, colliding with an identical mass, m, initially at rest.

2. a mass, 2m, colliding with a mass, m, initially at rest.

3. a mass, m, colliding with a mass, 2m, initially at rest.

1.  (m with v) collides with (m initially at rest).

mvb = 2mva              vb = 2va   

K = (1/2)*m*(vb)^2 

ΔK = (1/2)*2m*(va)^2- (1/2)*m*(vb)^2

ΔK/K= -50%      

2. (2m with v) colliding with (m initially at rest).

mvb = 3mva              vb = 3va

 K = (1/2)*m*(vb)^2

 ΔK = (1/2)*3m*(va)^2- (1/2)*m*(vb)^2

          

 ΔK/K= -66.7%

3. (m with v) colliding with (2m initially at rest).

   2mvb = 3mva        vb = (3/2)va

 K = (1/2)*2m*(vb)^2

 ΔK = (1/2)*3m*(va)^2- (1/2)*2m*(vb)^2

         

 ΔK/K= -33.3%

Conclusion

From this lab I learned that momentum is conserved. In the inelastic collision the two carts with different masses and velocities impact each other, stick together, then become one mass with one velocity.  We did a run where a mass with a velocity impacted a mass that was at rest, a run where a twice the mass with a same velocity crashes into a mass that is at rest, a run where a mass with a velocity crashed into a cart with twice the mass. In every trail momentum was conserved. The law is somewhat consistent. As we started adding weights to the carts we the difference became less. The increase of weight had a dampening effect on the results of this experiment. Our overall average was 8.8% difference, which shows that momentum is mostly conserved. Something that could have affected it were air drag, friction, and the table not being perfectly level, the carts also stuck together by velcro patches which couldve contributed to our percent error.

Hooke’s Law and the Simple Harmonic Motion of a Spring

Introduction


To determine the force constant of a spring and to study the motion of a spring and mass
when vibrating under influence of gravity. When a spring is stretched a distance x from its equilibrium position, it will exert a restoring force directly proportional to this distance. We write this restoring force, F, as: F  =  -kx. where k is the spring constant and depends on the stiffness of the spring.  The minus
sign remind us that the direction of the force is opposite to the displacement.  Equation 1 is valid for most springs and is called Hooke’s Law. If a mass is attached to a spring that is hung vertically, and the mass is pulled down and released, the spring and the mass will oscillate about the original point of equilibrium. Using Newton’s second law and some calculus we can show that the motion is periodic (repeats itself over and over) and has period, T (in sec), given by  T =[(4pi^2)/k)m]^(1/2)

Calculations

 F  =  -kx


 T =[(4pi^2)/k)m]^(1/2)

Data

Mass (kg)	Force (N)	Initial (m)	Final (m)	Period (s)	Amplitude (m)	Frequency (Hz)	Period^2 (s)
0		0.177	0.277
0.35	-3.43	0.325	0.425	0.784	0.0895	1.276	0.5824
0.45	-4.41	0.369	0.469	0.884	0.0892	1.131	0.7488
0.55	-5.39	0.412	0.512	0.972	0.0857	1.029	0.9152
0.65	-6.37	0.452	0.552	1.056	0.1025	0.947	1.0816
0.75	-7.35	0.493	0.593	1.134	0.0891	0.882	1.2481
0.85	-8.33	0.538	0.638	1.21	0.0676	0.826	1.4145
0.95	-9.31	0.579	0.679	1.268	0.0887	0.789	1.5809
1.05	-10.29	0.618	0.718	1.346	0.0784	0.743	1.7473

 1050g Positon vs Time

Here is the postion vs time graph for the spring loaded with a mass of 1050g  with a sinusoidal fit

 350g Position vs Time

Here is the postion vs time graph for the spring loaded with a mass of 1050g 

Force vs Displacement of all Runs

This the Force vs Displacement graph for all of the runs 350g through 1050g

Conclusions
In this lab we went over Hooks law of springs. We learned how to find a spring constant (k) by having a known mass, a length of displacement of the spring. The spring is stretched and the mass is attached and let go for 10 seconds. The spring oscillates and the position vs time is recorded. the spring constant can be calculated using F=-kx

Ballistic Pendulum

Introduction:

In this lab we had to use the ballistic pendulum to determine the initial velocity of a projectile using conservation of momentum and conservation of energy.

To study this we use  the equation KE=PE. Kinetic energy equals the Potential Energy so energy is conserved.

 KE=PE

1/2 (M + m)(V^2) = (M+m)(g)(h)

To solve for velocity of the pendulum arm, we rearrange the equation to

V=(2gh)^(1/2)

We know that momentum is conserved so 

o

P(initial) = P(final)

or 

(M+m)((2gh)^(1/2))=mvo

Where v₀ is the initial velocity of the ball before impact. By using equations and we can find the inital velocity, of the ball.

Vo=[(M+m)(2gh^(1/2))]/m

This combines both the conservation of momentum equation and the conservation of energy equation and shows how we can find v0, which is the initial velocity of the ball 

We could also find the initial velocity of the ball if we just shoot it into the open from a known height.

so we use Kinematics to find the initial velocity 

X=Vox*t+(1/2)ax*t^2 =Vox*t+(1/2)0*t^2 = Vo*t

Y= Voy*t+(1/2)ay*t^2 = 0*t+(1/2)(-9.8m/s^2) = -4.9t^2

t = [Y/(-4.9)]^(1/2)

X=Vo*t

X=Vox*[Y/(-4.9)]^(1/2)

Vox=X/([Y/(-4.9)]^(1/2)

Data

We ran 9 trials for the ball on the pendulum and obtained the following heights in notches. Our average was 8.56 notches.

Trials	Distance (notches)
1	10
2	7
3	8
4	11
5	10
6	9
7	8
8	8
9	6
Average	8.56

Then to find the initial velocity we use

V=(2gh)^(1/2)
V=(2(-9.8m/s^2)(0.096))^(1/2)
V=1.37m/s

Vo=[(M+m)(2gh^(1/2))]/m

Vo=[(0.1994m+0.0573m)(1.37m/s)]/)0.0573m

Vo=6.137m/s

Then we shot the ball out into the open and recorded the distances it traveled

Trials	Distances X(m)
1	2.615
2	2.629
3	2.675
4	2.722
5	2.747
6	2.777
7	2.786
8	2.818
9	2.818
Average	2.7319

To calculate this we used 

Vox=X/([Y/(-4.9)]^(1/2)

Vox=2.73m/([1.006/(-4.9)]^(1/2)

Vox= 6.031

To find the percent error between the two values we did the following

{(V1-V2)/[(V1+V2)/2]}*100

{(6.031-6.137)/[(6.031+6.137)/2]}*100

(0.01742*100)

1.742% error

Conclusion

In conclusion, I learned that you can find the initial velocity of an object by combining the laws of conservation of momentum and the laws of conservation of energy. Energy is never lost , it just gets transformed into a different type of energy. We could find the initial velocity from the kinematics equations by rearranging them. We also found the initial velocity with energy equations. Our percent difference was 1.742% which is relatively low. Some sources of error could be measurement. We measured distances by stacking meter sticks. The friction of the notches was not taken into account and could have lowered the pendulums maximum height. Another friction that was not taken into account was the the part where the pendulum was attached to the stand. This was assumed to be frictionless. 

Human Power

Introduction

Power can be described as the rate at which work is done which can also be translated as the rate at which energy is converted from one from to another. To show this we look at the equation:

Change in PE = mgh

PE=potential energy

M=mass

G=gravity

H=height

We can use this equation to find the change in potential energy which we need because the equation for  power output is:

Power = (change in PE) / (change in time)

Unit Analysis

Potential Energy= mgh

PE=kg(m/s^2)(M)

PE=(kg(m^2))/s^2=J

Power= PE/t

Power= J/s

Watt=J/s

1 watt= 0.00134102209 horsepower

In this lab we had to run up a flight of stairs that was 4.29 m high and record the time it took to make it from the bottom of the stairs to the top of the stairs.

Power Output

Trial	Weight Force (N)	Height of Stairs (m)	Change in Potential Energy (J)
1	888.86	4.29	3813.21
2	888.86	4.29	3813.21
Trial	Change in Time (s)	Power Output (watts)	Average Power Output (watts)
1	13.6	280.38	399.95
2	7.34	519.51

Then to find my horsepower out I did the following:

399.95 watts*(0.00134102209 hp/1 watt)= 0.54 hp

Discussion

To find my percent difference in relation to the class average I did the following. 

% Difference= [(Measured Value of Class - My value)/ 0.5(class+me)]*100

Watts [(634.98-399.5)/0.5(634.98+399.95)]*100

[.45419]*100

45.4 % difference

Horsepower.... [(.85-.54)/0.5(.85+.54)]*100

[.44604]*100

44.6 % difference

The values that I obtained let me know that I had less power out put than the class The percent difference for watts was 45.4% and the percent difference for horsepower was 44.6%.  I put out roughly less then 50% of what the class put out. 

Questions

1.  Is it okay to use your hands and arms on the hand-railing to assist you in your climb up the 

stairs?  Explain why or why not

It is okay to use your arms to assist in the climb up the stairs because the force that is exerted on the hand rail is to push you up. That force comes from your body so there is no difference in hand force and leg force they can both propel you upward and thus change your gravitational potential energy.

2.  Discuss some of the problems with the accuracy of this experiment.

Some sources of inaccuracy in this experiment are measurement of the height of the stairs, Timing of the runs up the stairs, human reaction time, when the recorders decided to stop the timer, the moment it took to turn around in the stairs to head the other direction. Another is that we are not perfectly at sea level

Follow up Questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expends the most power? Explain your answer.

Hinrick does the same amount of work as Valdis in less time, Hendrick has higher power output.

2. A box that weighs 1000 Newtons is lifted a distance of 20.0 meters straight up by a rope and pulley system. The work is done in 10.0 seconds. What is the power developed in watts (w) and kilowatts (kw)?

Work = ΔPE = 1000 (N) * 20 (m) * 9.8(m/s2)= 196000 (J)

Power = ΔPE/Δt = 20000 (J) / 10 (s) = 2000 (w) or 2 (kw)

3.  Brynhildur climbs up a ladder to a height of 5.0 meters. If she is 64 kg:

a. What work does she do?

The work she does is to  pull herself up and increase height.

b. What is the increase in the gravitational potential energy of the person at this height?

W= ΔPE= mgh = 64 (kg) * 9.8 (m/s2)  * 20 (m) = 3136 (J)

c. Where does the energy come from to cause this increase in P.E.?

From her own arms and legs - she uses her arms to pull herself up and her legs to push herself up.

4. Which require more work: lifting a 50 kg box vertically for a distance of 2 m, or lifting a 25 kg box vertically for a distance of 4 meters?

Since work is defined as force x distance and both would result in the same amount,

50 kg * 9.8 (m/s2) * 2 (m) = 980 J

25 kg x 9.8 (m/s2) x 4 (m) = 980 J

Both of them do the same amount of work

Conclusions

In this lab I learned the that ∆ PE = mgh where m is mass g is gravity (9.8 m/s^2) and h is the height. To see how much power you put out when your height is changed you use Power  = ∆ PE/∆t , the change in potential energy is divided by the change in time. So power is put out by us climbing  the stairs but that power is not lost, the energy is converted from out kinetic energy running up the stairs to our change into our gravitational potential energy from now being at a different height. So our body used energy from our food to move us up (kinetic) and as we climbed the stairs we lost kinetic energy and gained Potential energy. So energy is conserved in every action, energy does not just disappear it is just transformed into a different type of energy.