Fall 07 Midterm 3 Solutions Ch. 6-8

Fall 07 Midterm 3 Solutions Ch. 6-8 - For all parts of this...

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A For all parts of this exam, use g = 9.80 m/s 2 as the gravitational acceleration at the surface of the Earth. 1. Scalar Product and Center of Mass (8 pts) a. Find the angle between the vectors A = 3.00 i – 2.00 j + 2.00 k and B = 1.00 i + 4.00 j – 3.00 k . You may express the angle in degrees or radians, but be sure to indicate which. (4 pts) A B = |A | | B | cos θ AB cos θ AB = A B /( |A | | B |) = (A x B x + A y B y + A z B z )/(A x 2 +A y 2 +A z 2 ) 1/2 / (B x 2 +B y 2 +B z 2 ) 1/2 A B = -11 (Version A), -13 (Version B), -15 (Version C), -17 (Version D) | A | = (9 + 4 + 4) 1/2 = (17) 1/2 = 4.1231 (all versions) | B | = (26) 1/2 = 5.0990 (Version A) cos θ AB = -11/(4.1231 × 5.0990) = -0.5232 θ AB = 122 ° or 2.12 rad b. A system of three objects is arranged as shown in the table. Find the (x,y) position of the system’s center of mass. (4 pts) Mass (kg) x Position (m) y Position (m) Object 1 8.00 -1.00 +1.75 Object 2 3.00 +2.50 +3.00 Object 3 5.00 -1.25 -1.50 x cm = (m 1 x 1 + m 2 x 2 + m 3 x 3 )/(m 1 + m 2 + m 3 ) x cm = (-8.00 kg m + 7.50 kg m – 6.25 kg m)/(16.0 kg) = (-6.75 kg m)/(16.0 kg) x cm = -0.422 m (all versions) y cm = (m 1 y 1 + m 2 y 2 + m 3 y 3 )/(m 1 + m 2 + m 3 ) y cm = (14.0 kg m + 9.00 kg m – 7.50 kg m)/(16.0 kg) = (15.5 kg m)/(16.0 kg) y cm = +0.969 m (Version A) 122 ° or 2.12 rad ( 0.422 m, +0.969 m)
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A 2. Open Systems and Energy Transfer (8 pts) a. A constant force is applied to a 4.50 kg object which starts at rest. After the object has moved 2.00 m from its initial position, its speed is 1.50 m/s. Find the magnitude of the force. (4 pts) Δ K = F Δ x Δ K = ½ m v f 2 – ½ m v 0 2 = ½ m v f 2 = F Δ x F = (m v f 2 )/(2 Δ x) = (4.50 kg)(1.50 m/s) 2 /(2 × 2.00 m) = 2.53 N (Version A) b. The rate of water flowing over Niagara Falls is 5720 m 3 /s. The water falls a distance 51.0 m. Assuming the density of water is 1000 kg/m 3 , what is rate (in Watts) at which gravitational potential energy is released? (4 pts) Power = (5720 m 3 /s)(1000 kg/m 3 )(9.80 m/s 2 )(51.0 m) = 2.86 × 10 9 W. 2.53 N 2.86 GW
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A 3. Closed Systems (8 pts) a. Briefly explain the distinction between conservative and non-conservative forces, giving an example of each type. How can we determine whether a force is conservative or not? (4 pts) A conservative force is one for which the work done by the force depends only on the starting and ending points of a displacement, not on the path taken in between. A conservative force can be described in terms of potential energy, while a non- conservative force cannot. (2 pts for giving a substantially correct definition) Examples of conservative forces are gravity, electromagnetism and the elastic force of a spring. Examples of non-conservative forces are friction and air resistance. (1 pt for giving a correct example of each type of force) Note it is NOT correct to say that a non-conservative force does not conserve energy. b.
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Fall 07 Midterm 3 Solutions Ch. 6-8 - For all parts of this...

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