• Support PF! Buy your school textbooks, materials and every day products Here!

Conservation of Momentum Lab

  • #1
So, here is my data:
TRIAL 1 TRIAL 2 TRIAL 3
MASS OF CAR 1 7.5 N 7.5 N 7.5 N
MASS OF CAR 2 7.5 N 7.5 N 7.5 N
TIME FOR CAR 1 (s) 8.63 8.31 9.40
TIME FOR CAR 2 (s) 7.50 10.72 8.78
DISPLACEMENT FOR CAR 1 (m) 305.00 365.00 348.60
DISPLACEMENT FOR CAR 2 (m) 387.50 369.10 305.10


And the questions from the lab are:
1. Find the velocity of each car after the collision. Remember it is
accelerating from rest.
2. Find the momentum for each car before and after the collision. Then find
the total momentum before the collision and the total momentum after
the collision. p1 + p2 = p1+ p2
3. Determine the % difference for the four different situations.

(The lab consisted of two cars placed side by side, when a button was pushed on one car a rod would extend and cause a collison, I hope that made sense!)



My issues:
-I kind of get the feeling from the first question that I may have used an incorrect approach in solving this, but I just divided the displacement by the time. (ex. 385.50m/7.50s=51.7m/s)
-Okay, for the second question, I would use p=mv to find the momentum. But wouldn't the momentum before the collision always be zero since it begins from rest? so wouldn't it end up looking like 0 = (whatever the added momentums after the collisons are)? meaning momentum is not conserved? I thought momentum was always conserved but maybe I am completely wrong.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
The cars move in opposite directions don't they?
 

Related Threads for: Conservation of Momentum Lab

Replies
5
Views
4K
Replies
3
Views
4K
Replies
8
Views
5K
Replies
8
Views
816
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
11
Views
2K
Top