Physics, Please help!?
Reminder: No collision is truly elastic!
In procedure 3: assume cart 1 has mass 422 g and initial velocity +v10 = 3.98 m/s, and cart 2 has mass 836 g. Assume the track is frictionless, and after the "elastic" collision cart 1 is moving at v1 = -1.277 m/s and cart 2 is moving at v2 = 2.582 m/s.
a) What percentage of the original energy was lost in this collision?
b)What percentage of the original linear momentum was "lost" due to external forces? (Remember: linear momentum is a vector!)
- NCSLv 74 years agoFavorite Answer
First conserve momentum to find the initial velocity of cart 2:
422g*3.98m/s + 836g*V = 422g*-1.277m/s + 836g*2.582m/s
Solve for V; it will be negative.
a) Find the initial KE of the two carts
and the final KE of the two carts.
KE = Σ ½mv²
%energy lost is (KEinitial - KEfinal) * 100% / KEinitial
b) total the LHS and then the RHS of the momentum equation.
(LHS - RHS) * 100% / LHS
is the percentage of momentum lost.
Hope this helps!