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solution2ExamBSP08

# solution2ExamBSP08 - Answer key for second exam version B 1...

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Answer key for second exam version B 1 1. This problem is solved in three steps. When the hockey puck leaves the top of the cliff, it is traveling horizontally with a speed v 1 . It will then be a projectile falling to the ground from an initial height y 0 = 8 . 5 meters and with vertical acceleration g . This projectile will also travel at constant horizontal speed v 1 . We can find v 1 by computing the fall time t , and then divided the horizontal distance traveled (6.2 meters) by that time t . Then we can use conservation of energy to find the energy at the top of the cliff, which must be all kinetic energy when the puck started with speed v 0 . Falling body with no initial speed y ( t ) = 0 = y 0 - 1 2 gt 2 Time to fall 8.5 meters = t = v u u t 2 y 0 g = v u u t 17 9 . 8 = 1 . 32 seconds Horizontal speed at the top of the cliff v 1 = 6 . 50 1 . 32 = 4 . 84 m/s Total Energy at the top of the cliff = mgh + 1 2 mv 2 1 Total Energy at the start = 1 2 mv 2 0 Conservation of energy = 1 2 mv 2 0 = mgh + 1 2 mv 2 1 = v 2 0 = 2 gh + v 2 1 = 2 * 9 . 8 * 8 . 5 + (4 . 84) 2 (m/s) 2 = v 0 = 13 . 8 m/s Answer is D 2. First calculate how much total energy is added to the 2000 kg of sand in 60 seconds. The power will be that total energy divided by 60 seconds. The total energy is the potential energy increase in rising 12 meters plus the kinetic energy in acquiring a speed of 5 m/s. E total = Mgh + 1 2 Mv 2 = 2000(9 . 8 × 12 + 0 . 5 × (5) 2 ) = 2 . 602 × 10 5 Joules P = E t = 2 . 602 × 10 5 Joules 60 seconds = 4340 Watts Answer is C

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Answer key for second exam version B 2 3. This is a static equilibrium situation where the net forces on each of the two objects A and B are both zero. For object A that means the tension in the connecting rope is equal to the weight of A . That tension T = w A acts to the right on object B . The force Q acts to the left on B , attempting to move B to the left. That means there is a static friction force f max s which is at its maximum acting to the right on B , preventing B from moving to the left. In turn, the static friction force is given by the normal force of the table, which is the sum of the weight of B and the downward push force P which is acting vertically on B . f max s = μ s N = μ s ( w B + P ) Q = T + f max s = w A + μ s N = w A + μ s ( w B + P ) Q = 19 * 9 . 8 + 0 . 40 * (15 * 9 . 8 + 60) = 269 N Answer = B 4. The average force is found from the impulse equation using the net momentum change F Δ t = Δ p = m Δ v . In this case, taking the positive direction as the initial direction of the ball, Δ v = 11 - ( - 11) = 22 m/s. So F = (0 . 8) * (22) / 0 . 05 = 352 N. Answer = A 5. Doubling the compression of a spring means that it contains four times the amount of potential energy. When all of that potential energy is transferred to the mass, the mass will have four times the kinetic energy. Quadrupling the kinetic energy means that the speed has doubled. Answer = A 6. Simple plug-in solution from the banked curve speed equation v max = Rg tan θ = θ = tan - 1 ( v 2 max /Rg ) = tan - 1 ((78) 2 / (190 * 9 . 8)) θ = 73 . 0
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solution2ExamBSP08 - Answer key for second exam version B 1...

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