May 16, 2008
PHY2053 Discussion
Warm Up Quiz (Appendix A)
Name:
UFID:
*1. Determining an angle is essential to solve problems in physics 1. Suppose a person is
towing up a block on a 15˚ incline by pulling a string attached to the block. The string
makes 20˚ to the incline. Find the following angles (0˚≤θ≤90˚): (1). The angle the string
makes to the vertical. (2). The angle the string makes to the line perpendicular to the
incline.
The string makes an angle of 35˚ (= 15˚+20˚) to the horizontal. Thus, the angle the
string makes to the vertical is
90˚-35˚ = 55˚.
Since the string makes 20˚ to the surface of the incline, thus the angle between the
string and the line perpendicular to the surface is
90˚-70˚ = 20˚
***2. When you throw a ball, the height of the ball is given by a quadratic function of
time and the horizontal distance is given by a linear function of time. Thus you need
solve a quadratic equation and a linear equation simultaneously. Solve the following
pair of simultaneous equations: x-20cos45˚t = 0, tan30˚x+20sin45˚t-4.9t²= 0. (The
solutions give the time of flight and horizontal range of a projectile thrown at an angle
of 45˚ on a slope of -30˚.)
We solve the first equation for x and plug it in the second equation, and then solve the
equation for t:
x = 20cos45˚t
tan30˚×20cos45˚t +20sin45˚t-4.9t²= t(tan30˚×20cos45˚ +20sin45˚-4.9t) = 0
⇒ t = 0, (tan30˚×20cos45˚ +20sin45˚)/4.9 = 4.55
Substituting these values in the first equation, we have
x = 0, 20cos45˚×4.55 = 64.3
Thus if you throw a ball at t = 0 from the origin, it hits the ground in 4.55 s and the
horizontal range is 64.3 m.
**3. Some problems in physics 1 require you to solve quartic equations. However, all the
quartic equations appear in physics 1 have a form of Ax⁴+Bx²+C = 0. So letting x² = y ,
you can solve them as quadratic equations. Solve 8cos⁴θ-2cos²θ-3 = 0 and determine all
the solution for 0˚≤θ≤180˚.
Letting cos²θ be x, we solve the quadratic equation for x:
8x²-2x-3 = 0 ⇒ x = (2±√(2²+4×8×3))/(2×8)= 3/4, -1/2
Since cos²θ is positive (The square of any real number is positive.), x = 3/4.
Solving cos²θ = 3/4 for θ, we have
cosθ = ±√(3)/2 ⇒ θ = 30˚, 150˚
*4. The sensation of loudness is known to be proportional to the intensity of sound (You
don’t need to know what the intensity is yet.), and it is measured as the decibel level,
which is defined by β(dB) = 10log₁₀(I/I₀), where I is the intensity of the sound and I₀
is called the reference intensity. Now two sounds have intensities of I₁ = 3.00×10⁸I₀
and I₂ = 1.00×10⁸I₀. By how many decibels (dB) is the decibel level of sound 1 higher
than that of sound 2?
All you have to do is to plug the given numbers in the definition of decibel level and
calculate the difference:
β₁-β₂ = 10log(3.00×10⁸)-10log(1.00×10⁸)
= 10log(3.00×10⁸/1.00×10⁸) = 10log(3) = 4.77 dB
, where I used log(A)-log(B) = log(A/B).