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Unformatted text preview: Answer key for second exam version A 1 1. 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 = M gh + 1 2 M v 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 A 2. A ball which is rising is still being affected by the force of gravity, which is external to the ball. So its momentum is not conserved. In fact, its momentum is continually decreasing since its speed is continually decreasing as it rises. Its potential energy and kinetic energy are also continuously changing, but the mechanical energy sum remains constant. Answer is A 3. If the same force acts over twice the distance then the kinetic energy change will be twice as much. Since K ≡ mv 2 / 2, a doubling of the kinetic energy means that the speed has increased by √ 2. Answer is D 4. First we need to get the ( x, y ) coordinates of each mass 1) 2 kg mass at (2 , 0) 2) 6 kg mass at (5 , 2) 3) 4 kg mass at (3 , 5) 4) 8 kg mass at (0 , 3) Now we can compute x cm and y cm according to the sums over positions weighted by the masses: x cm = ∑ 4 i =1 m i x i ∑ 4 i =1 m i = 2 * 2 + 6 * 5 + 4 * 3 + 8 * 2 + 6 + 4 + 8 = 2 . 3 y cm = ∑ 4 i =1 m i y i ∑ 4 i =1 m i = 2 * 0 + 6 * 2 + 4 * 5 + 8 * 3 2 + 6 + 4 + 8 = 2 . 8 Answer is A Answer key for second exam version A 2 5. 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 = 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 . Falling body with no initial speed y ( t ) = 0 = y 1 2 gt 2 Time to fall 8.5 meters = ⇒ t = v u u t 2 y 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 Conservation of energy = 1 2 mv 2 = mgh + 1 2 mv 2 1 = ⇒ v 2 = 2 gh + v 2 1 = 2 * 9 . 8 * 8 . 5 + (4 . 84) 2 (m/s) 2 = ⇒ v = 13 . 8 m/s Answer is D 6. Conservation of energy requires that at the highest point of the motion, where the speeds and the kinetic energies are zero, then the potential energies must equal the initial kinetic energies. The initial potential energy can be taken as zero. Ball B with twice the speed of ball A then must travel four times as high as ball A since it has four times the kinetic energy of ball...
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 Spring '08
 Maguire
 Physics, Energy, Force, Friction, Kinetic Energy, Potential Energy, Power, hockey, wA

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