06 Final Solutions

# 06 Final Solutions - A 1 Vectors and Units a An object is...

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Unformatted text preview: A 1. Vectors and Units a. An object is displaced 5.00 m in a direction 30.0° north of east, then 3.00 m in a direction 50.0° west of north. Find the magnitude of the resultant displacement. Call the first displacement Δ r 1 and the second Δ r 2 . Δ r 1 = (5.00 m) cos 30 i + (5.00 m) sin 30 j = 4.330 m i + 2.500 m j Δ r 2 = (3.00 m) cos 140 i + (3.00 m) sin 140 j = -2.298 m i + 1.928 m j Δ r = Δ r 1 + Δ r 2 = (2.032 m) i + (4.428 m) j | Δ r | = [ (2.032 m) 2 + (4.428 m) 2 ] 1/2 = 4.87 m b. The power carried by a wave traveling on a string is given by the expression: P = ½ μ ω 2 A 2 v, where A is an amplitude of the wave (dimensions of length, L), ω is the wave’s angular frequency (dimensions of inverse time, 1/T) and v is the speed of the wave. Find the dimensions of μ and express them in terms of mass (M), length (L) and time (T). Power has dimensions of energy over time: P = E/T Energy has dimensions of force times distance: E = F L, so P = F L/T Force has dimensions of mass times acceleration: F = M L/T 2 , so P = M L 2 /T 3 That makes the expression (dimensionally): M L 2 /T 3 = μ (1/T 2 ) L 2 (L/T) = μ L 3 /T 3 Solving for the dimensions of μ : μ = M/L ( μ is the mass per unit length of the string) M/L 4.87 m A 2. Velocity and Acceleration a. An object’s position at time t is given by x(t) = 3.00 m – (2.00 m/s) t + (1.00 m/s 2 ) t 2 Find the object’s velocity at time t = 4.00 s. v(t) = dx/dt = -2.00 m/s + 2(1.00 m/s 2 ) t v(4.00 s) = 6.00 m/s b. Starting from rest and moving in a straight line, an object accelerates to 90.0 m/s in 10.0 s, and then decelerates at -7.50 m/s 2 until it returns to rest. Find the object’s total displacement from the starting position. During the first 10.0 s, the displacement is Δ x 1 = ½ (v i + v f ) t = (45.0 m/s) (10.0 s) = 450.0 m During the second interval, the displacement is given by: v f 2 – v i 2 = 2 a Δ x 2- (90.0 m/s) 2 = 2 (-7.50 m/s 2 ) Δ x 2 Δ x 2 = 540 m Total displacement Δ x = Δ x 1 + Δ x 2 = 990 m. (There are other ways to do the problem, of course) 990 m 6.00 m/s A 3. Two-dimensional Motion Ignore air resistance for both parts of this problem. a. A ball lying on the ground is kicked into the air and lands 3.50 s later, 50.0 m downfield from its initial location. With what speed was the ball kicked? Initial horizontal velocity: v x0 = Δ x/ Δ t = (50.0 m)/(3.50 s) = 14.286 m/s Initial vertical velocity: v y0 There are several possible ways to find this. The easiest to recognize that the ball reaches it highest point (v y = 0) at time Δ t/2 = 1.75 s Then Δ v = 0 – v y0 = a ( Δ t/2) = -g Δ t/2 v y0 = g Δ t/2 = (9.80 m/s 2 ) (1.75 s) = 17.150 m/s The speed v = (v x0 2 + v y0 2 ) 1/2 = 22.3 m/s b. A 0.250 kg object on a 0.500 m tether is swung in a vertical circle. The object’s speed at the bottom of the circle is 8.50 m/s. Find the tension in the rope when the object is at the top of the circle. Note that the object’s speed is not constant,...
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06 Final Solutions - A 1 Vectors and Units a An object is...

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