Physics Post Lab 5 - 3. The mud ball transferred more...

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Annie Markowitz Post Lab 5 1. If this was a semi truck hitting a marble, the mass of the marble would be negligible, so: v(f) large glider = [(m large glider – m small glider ) / (m large glider + m small glider )] * v(i) large glider v(f) large glider = (m large glider ) / (m large glider )* v(i) large glider v(i) large glider v(f) small glider = [2m large glider / (m large glider + m small glider )] * v(i) large glider v(f) small glider = (2m large glider / m large glider ) * v(i) large glider 2v(i) large glider 2. It’s better to use the first two times because in an inelastic collision, the third time occurs after the two carts were already stuck together. In order to complete the calculations accurately, you need gate times of the carts individually before the collision, because they have different properties once they collide.
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Unformatted text preview: 3. The mud ball transferred more momentum to the wall because they share the total momentum in the end since its an inelastic collision. The rubber ball has the total momentum and the wall has none because its an elastic collision. 4. Elastic collision pushing Cart 2B into stationary Cart 1. v(f) 2B = [(m 2B m 1 ) / (m 2B + m 1 )] * v(i) 2B will be negative because m 1 > m 2B v(f) 1 = 2m 2B / (m 2B + m 1 ) * v(i) 2B will be positive 5. m 1 v(i) 1 + m 2 v(i) 2 = (m 1 + m 2 )v(f) (3300lbs)(0mph) + (1500lbs)(v(i) 2 ) = (3300lbs + 1500lbs)(17.295mph) v(i) 2 = 55.344mph 6. 4 times more energy. KE = 1/2mv 2 KE(60mph) = (60 2 ) = 1800 KE(30mph) = (30 2 ) = 450...
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This note was uploaded on 11/29/2011 for the course PHY 102M taught by Professor Staff during the Fall '08 term at University of Texas at Austin.

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