Quizzes 


Quizzes given to Sections 10, 11, and 12 (PHYS 101) 042 Semester 
Chapter 1 Problem A A barrel has a radius of 20.0 in, a height of 25.0 in, and a mass of 600.0 mg. What is its total mass (in kg) if it is filled with benzene? The density of benzene is 0.87873 gm/cm^{3}? Solution: Volume of cylinder = π r^{2 }h = 3.14 (20.0 in)^{2} (25.0 in) (2.54 cm/1 in)^{3} = 515 x 10 3 cm^{3} Mass =density x volume = (0.87873 g/cm^{3}) (1 kg/1000 g) (515 x 10^{3} cm^{3}) = 452.15 kg Therefore, total mass = 452.15 kg + 0.6 kg = 452.75 kg ~ 453 kg Problem B What is the mass (in kg) of a lead semisphere of radius 5.0 in? (Density of lead is 11.342 g/cm^{3}) Solution: Volume of semisphere = (0.5) (4/3) π r^{3} = (2/3) (3.14) (5.0 in)^{3} (2.54 cm/1 in)^{3} = 4287.948 cm^{3} Mass = density x volume = (11.341 gm/cm^{3}) (1 kg/ 1000 g) (4287.948 cm^{3}) = 48.63 kg ~ 49 kg.
Chapter 2 Problem 1 The position of a particle in meters is described as x(t) = 6t^{2}  4t +3 (a) What is v_{avg} in the period from t = 0 s to t = 5 s? (b) What is v at t = 3 s? (c) What is the acceleration when the particle stops momentarily? Solution: (a) v_{avg} = (Δx/Δt) = [x(t=5)  x(t=0)] / (50) = (130 m) / (5 s) = 26 m/s (b) v = dx/dt = 12 t  4 v at (t = 3 s) is 32 m/s (c ) The particle stops momentarily when v = 0, i.e., 12 t  4 = 0 gives t = 0.33 s However, we find from the equation (a = dv/dt = 12 m/s^{2}) that the acceleration is always constant (at every moment). Problem 2 A) A stone is thrown upward from the top of a building with an initial speed of 1 m/s. If the building is 20m high what is the velocity of the stone when it his the bottom? B) Solve it also when the stone is thrown downward with initial velocity 1 m/s? Solution: A) Initial conditions: v_{0} = +1 m/s a =  g =  9.8 m/s^{2} y_{0} = 0 m (if we take the origin coordinate where the motion starts) y =  20 m v^{2} = v_{0}^{2}—2g(y– y_{0}) = (1 m/s)^{2}—2 (9.8 m/s^{2})(20 m) v = 19.8 m/s downward B) Initial conditions: v_{0} = 1 m/s a =  g =  9.8 m/s^{2} y_{0} = 0 m (if we take the origin coordinate where the motion starts) y =  20 m v^{2} = v_{0}^{2}—2g(y– y_{0}) = (1 m/s)^{2}—2 (9.8 m/s^{2})(20 m) v = 19.8 m/s downward ( same answer)
