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# 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!)

### 1 Answer

- 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!

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