An enthusiastic group of students are busy doing experiments. This time they are
ID: 2141079 • Letter: A
Question
An enthusiastic group of students are busy doing experiments. This time they are exploringelastic and inelastic collisions, much like you will be doing in the upcoming lab.
First, the students consider an elastic collision. The students set up a cart with a spring attached to the force probe much like the setup in Lab 7s Activity 3-2. They weigh the cart assembly and find that its mass is 732.4 g. They launch the cart towards the target and collect the following force-vs.-time data [NOTE: "ms" means "milli-seconds"]:
NOTE: This plot shows actual data collected with a setup as described above.
t [ms] F [N] 0.0 0.029 17.1 0.159 34.2 0.253 51.3 0.396 68.4 0.503 85.5 0.647 102.6 0.745 119.7 0.863 136.8 1.062 153.9 1.103 171.0 1.204 188.1 1.381 205.2 1.497 222.3 1.596 239.4 1.682 256.5 1.751 273.6 1.858 290.7 1.964 307.8 2.092 324.9 2.182 342.0 2.290NOTE: This plot shows actual data collected with a setup as described above.
Explanation / Answer
For part a) we need to take ?t, and multiply it by F.
- The distance between each time interval is 17.1 ms, because theanswer needs to be in N*s, this turns into .0171s
- ?t = .0171
- F is the sum of all forces, so once you add them all up, youget:
- F = 40.762
Multiply the two for your answer.
b)
To find the velocity, we need first the acceleration. You do thisby dividing each force by your given mass (Don't forget to convertto kg!)
-If you were to plot those acceleration points against time, thevelocity would be the area under the curve.
-Because you collision occurs exactly halfway through the points,so you want the area under the curve from t = 0 to t = 171ms
-Sum up all of those accelerations and divide by 2 (because you areonly looking for half).
-Take that number and multiply by ?t for your speed before thecollision.
c)
In a perfectly elastic collision, momentum is conserved. In thiscase the mass stays the same ? the velocity will be equal andopposite to the velocity before.
NOTE: The question does not ask for the velocity, rater thespeed.
d)
Found exactly like before, except because the object does notrebound, you need all of the area under the curve, not justhalf.
-So you take the impulse given, and divide it by ?t. Thisgives you the the area under the force curve.
-Divide that by the mass (in kg) given at the start of the problem.This will give you the area under the acceleration curve.
-Multiply that by ?t again to find your velocity.
e)
In a perfectly inelastic collision, Kinetic energy is notconserved. Basically, the first object sticks to the second object.In the this case the second object remains stationary so... howdoes the cart act?