06 Midterm 1 Solutions Ch. 1-3

06 Midterm 1 Solutions Ch. 1-3 - 1. Units and Vectors (8...

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Version A C B A 1. Units and Vectors (8 points) a. In atomic physics, the Bohr radius a 0 of the Hydrogen atom (describing the size of a Hydrogen atom in its ground state) is given by the expression: 0 2 e h a mc π α = , where m e is the mass of the electron, c is the speed of light, and is a dimensionless constant. The constant h is called Planck’s constant, and is one of a handful of truly fundamental physical constants. Find the dimensions of Planck’s constant, in terms of length (L), time (T) and mass (M). The dimensions of a 0 , for instance, would be “L”, since it is a length. L = h/(M L/T) = h T / (M L) h = M L 2 /T b. Suppose A + B + C = 0, where A , B , and C are two-dimensional vectors. You know that C = - C j , where j is the unit vector in the y-direction, and C is the magnitude of the vector C. You also know that A = -A cos θ i + A sin θ j and B = B sin θ i + B cos θ j , where θ is same angle in both cases (this means the vectors A and B are perpendicular to each other). Find B (the magnitude of B ) in terms of C and θ . (Hint: It may help to remember that cos 2 θ + sin 2 θ = 1) A x + B x + C x = 0 A y + B y + C y = 0 B x = B sin θ = - (C x + A x ) = - (-Acos θ ) = A cos θ So A = B tan θ B y = B cos θ = - (C y + A y ) = - ( - C + A sin θ ) = C - B tan θ sin θ So B (cos θ + tan θ sin θ ) = C Multiplying both sides by cos θ and applying the identity given in the hint gives: B = C cos θ M L 2 / T C cos θ
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Version A y x 2. Velocity and Speed (8 points) a. The pointer of a circular gauge is 3.00 cm long, and moves around the dial of the instrument at a constant rate of 1.25 radians/s. Find the speed of a point at the outer end of the pointer. The distance traveled is d = R θ , when θ is in radians. So v = R d θ /dt = (3.00 cm) (1.25 rad/s) v = 3.75 cm/s b. A baseball player hits a triple, and arrives at third base 12.5 seconds after leaving home plate. The sides of a baseball diamond are 27.4 meters in length. Find the player’s average velocity vector using a coordinate system where the y-axis points from home plate to the pitcher’s mound and second base. Let d = length of one side of diamond = 27.4 m Then the displacement Δ r = - d cos 45° i + d sin 45° j Δ r = - 0.707 d i + 0.707 d j The average velocity v = Δ r / Δ t v = - (0.707) (27.4 m/12.5 s) i + (0.707) (27.4 m/12.5 s) j v = - 1.55 i + 1.55 j m/s 3.75 cm/s -1.55 i + 1.55 j m/s
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Version A 3. Acceleration (8 points) a. An object, starting from rest and moving along a straight line, accelerates at +2.25 m/s 2 for a time t, and then decelerates at -1.75 m/s 2 for the same time t. The object’s total displacement is 125 meters. Find the time t. Δ x 1 = ½ a 1 t 2 (object starts from rest, so v 0 = 0) After time t, the object’s velocity will be v 2 = a 1 t (initial velocity for second interval) Δ x 2 = ½ a 2 t 2 + v 2 t = ½ a 2 t 2 + (a 1 t)t = (½ a 2 + a 1 ) t 2 Δ x 1 + Δ x 2 = d = 125 m = ½ a 1 t 2 + (½ a 2 + a 1 ) t 2 = (a 2 + 3 a 1 )/2 t 2 t 2 = 2d/(3 a 1 + a 2 ) () 22 12 2 125 m 2 7.07 s 3 3 2.25 m/s 1.75 m/s d t aa == = + ⎡⎤ ⎣⎦ b. An object, starting from rest and moving along a straight line, travels +40.0 m with constant acceleration a, and then travels an additional 25.0 m in the same direction with constant acceleration – a . The object’s final velocity is +30.0 m/s.
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06 Midterm 1 Solutions Ch. 1-3 - 1. Units and Vectors (8...

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