2.) Our intrepid group of 142W students is at it again.This time they are explor
ID: 1742123 • Letter: 2
Question
2.) Our intrepid group of 142W students is at it again.This time they are exploring elastic andinelastic collisions, much like you will be doing in theupcoming lab.
First, the students consider an elastic collision. Thestudents set up a cart with a spring attached to the force probemuch like the setup in Lab 7’s Activity 2-2. Theyweigh the cart assembly and find that its mass is 838.3 g. They launch the cart towards the targetand collect the following force-vs.-time data [NOTE:“ms” means “milli-seconds”]:
NOTE: This plot shows actual data collected with a setup asdescribed above.
a) What is the impulse (J) imparted tothe cart? Recall (see, for example, the discussion inActivity 1-2) that the impulse is given by the area under theforce-vs.-time curve. For the purposes of this question,approximate this area by adding up the individualF·t (t being thetime between successive measurements) areas. This technique is asimple (yet quite good in this case) “numericalintegration”.
2 N·s
b) Assuming a perfectly elastic collision, what is yourbest estimate of the speed of the cart before the collision?
3 m/s
c) Assuming a perfectly elastic collision, what is yourbest estimate of the speed of the cart after the collision?
4 m/s
The students then explore inelastic collisions. They replace thespring with a dart so that the cart sticks to the target when ithits. They get the following data:
The students discover that Data Studio has an “area underthe curve” feature. They use this and find the total areaunder the curve to be 0.4542 N·s.
d) Assuming a perfectly inelastic collision, what is yourbest estimate of the speed of the cart before the collision?
5 m/s
e) Assuming a perfectly inelastic collision, what is yourbest estimate of the speed of the cart after the collision?
6 m/s
NOTE: This plot shows actual data collected with a setup asdescribed 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? Good luck on your lab, I hope its not as long as the last one ;)
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