# 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/cm3?

Solution:

Volume of cylinder = π r2 h

= 3.14 (20.0 in)2 (25.0 in) (2.54 cm/1 in)3 = 515 x 10 3 cm3

Mass =density x volume

= (0.87873 g/cm3) (1 kg/1000 g) (515 x 103 cm3) = 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 semi-sphere of radius 5.0 in? (Density of lead is 11.342 g/cm3)

Solution:

Volume of semi-sphere = (0.5) (4/3) π r3

= (2/3) (3.14) (5.0 in)3 (2.54 cm/1 in)3 = 4287.948 cm3

Mass = density x volume

= (11.341 gm/cm3) (1 kg/ 1000 g) (4287.948 cm3) = 48.63 kg ~ 49 kg.

Chapter 2

Problem 1

The position of a particle in meters is described as x(t) = 6t2 - 4t +3

(a) What is vavg 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) vavg = (Δx/Δt)

= [x(t=5) - x(t=0)] / (5-0) = (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/s2)

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 20-m 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: v0 = +1 m/s

a = - g = - 9.8 m/s2

y0 = 0 m (if we take the origin coordinate where the motion starts)

y = - 20 m

v2 = v02—2g(y– y0) = (1 m/s)2—2 (9.8 m/s2)(-20 m)

v   = 19.8 m/s downward

B) Initial conditions: v0 = -1 m/s

a = - g = - 9.8 m/s2

y0 = 0 m (if we take the origin coordinate where the motion starts)

y = - 20 m

v2 = v02—2g(y– y0)

= (-1 m/s)2—2 (9.8 m/s2)(-20 m)

v   = 19.8 m/s downward ( same answer)