Physics
NYA Mechanics
Examination

Part I: Problems

Solve all seven problems. Show all of your work, clearly and in order, to receive full marks. If you use a formula not given on the formula sheet, a derivation must be shown.

A car slows down from an initial velocity of 90.0 km/h to a final velocity of 36:0 km/h at a constant rate while undergoing a displacement of 105 m.

(a) What is the acceleration of the car?

(b) If the car continues with the same acceleration, how much additional time will it take to come to a stop?

(c) What is the total displacement of the car from the instant when it's travelling at 90km/h until it stops?

(d) Sketch the position-time curve for the car on the empty graph on the next page. Indicate the time and position at which it stops

Answers:


(a) Let's apply:

v² - vo² = 2 a ( x - xo)

a = (v² - vo²)/2 (x - xo)

v0 = 90km/h = 90 x 1000/3600 = 25 m/s v = 36 km/h = 36 x 1000/3600 = 10 m/s


a = (10² - 25²)/2 x 150 = -2.50 m/s2

a = - 2.50 m/s²

(b) Let's apply:

v = at + vo
t = (v - vo)/a = (0 - 10)/(- 2.5) = 4 s

t = 4 s

(c) Let's apply:

x = xo + vo t + (1/2) a t²

x = 105 + 10 t + (1/2) (- 2.25) t²

x = 105 + 10 t - (2.25/2) t²

At t = 4 s , x = 105 + 10 (4) - (2.5/2) 4² = 125 m

x = 125 m

(d) See the figure.

A projectile is launched at an angle of 53.1^o above the horizontal at a speed of 48:0m/s. The projectile lands on a 50.0 m high ledge.

(a) What is the horizontal distance from the launch point to the landing point ?
(b) What are the magnitude and direction of the projectile's velocity when it lands ?
(c) What is the maximum height reached by the projectile ?


Answers:

(a) y = yo + vo t sin θ - (1/2) g t²
With yo = 0 , we have:

y = vo t sin θ - (1/2) g t²

50 = vo t sin θ - (1/2) g t²

Δ = vo² sin² θ - 100 g =
48² sin² 53.1 - 100 g = 493.40

t₂ = (vo sin θ + √Δ)/g = 6.18 s

t₂ = 6.18 s

We have
x = t₂ vo cos θ
x = t₂ vo cos θ 6.18 x 48 x cos 53.1 = 178.20 m

x = 178.20 m

(b) v² = vx² + vy²

vy = vo sin θ - g t₂
vy ² = vo² sin² θ - 100 g

Then v² = vo² - 100 g
v = √(vo² - 100 g)
v = √(48² - 100 x 9.8) = 36.4 m/s

v = 36.4 m/s

α = cos^-1 (vo cos θ/v) =
= cos^-1 (48 cos 53.1/36.4) = 37.65^o

α = 37.65^o

(c)
2 g h_max = vo² sin² θ - 0
h_max = vo² sin² θ/2 g
h_max = 48² sin² 53.1 /2 x 9.8 = 75.2 m

h_max = 75.2 m

An 18.0 kg block is suspended from three cables as shown below. What are the tensions T1, T2, and T3?

Answers:

Along x: T2 cos 50 = T3 sin 30

T2 = T3 sin 30 /cos 50

Along y : T1 = T2 cos 40 + T3 cos 60

T1 = mg
T1 = 18 x 9.8 = 176.40 N

T1 = 176.40 N

T2 = mg/(cos 40 + cos 50/cos30)

T2 = 115 N

T3 = mg/(1 + cos 40 x cos 30 /cos 50)

T3 = 155 N

A 20.0 kg block on a horizontal surface is connected to a hanging 8.00 kg block by a string, as shown below.

(a) What is the minimum coefficient of static friction between the 20.0 kg block and the surface that would prevent the masses from moving ?
(b) If the surface is frictionless, what is the acceleration of the 20.0 kg block and the tension in the string ?

Answers:

(a)
μ_s 20 x g ≥ 8 x g
μ_s 20 ≥ 8/20 = 2/5 = 0.40

Minimum μ_s = 0.40

(b)
Along x : T = m1 a

Along y : m2 g - T = m2 a

Then:

a = m2 g
a = m2 g/(m1 + m2)

a = m2 g/(m1 + m2)

a = m2 g/(m1 + m2) = 8 x 9.8/28 = 2.8 m/s2

a = 2.8 m/s2

T = m1 a = 20 x 2.8 = 56 N

T = 56 N

A 1.60 kg block starts from rest and slides along the surface shown below. The surface is frictionless except for the rough patch indicated. After crossing the rough patch the block hits a spring.

(a) What is the speed of the block just before it enters the rough patch ?
(b) What is the speed of the block after crossing the rough patch ?
(c) How far does the block compress the spring before it comes to rest ?


Answers:

(a) (1/2) m vb² - 0 = mg h
vb = √(2 g h)
vb = √(2 x 9.8 x 1.5) = 5.42 m/s

vb = 5.42 m/s

(b) (1/2) m va² - (1/2) m vb² = μ m g x
vb² = va² - 2 μ g x
va² = 5.42 ² - 2 (0.3) (9.8) (3) = 29.38 - 17.64 = 11.74
va = 3.43 m/s

va = 3.43 m/s

(c)
(1/2) m va² = (1/2) k Δx²
m va² = k Δx²
Δx² = (m/k) va²
Δx = (m/k)^1/2 va
Δx = (1.6/600)^1/2 3.43 = 0.177 m

Δx = 0.17.7 cm

A stack of two transparent plates is surrounded by air. The indices of refraction of the plates are 1.50 and 2.18, respectively. Light in air is incident on the firrst plate at an angle of 60.0^o relative to the normal.

(a) What is the angle of refraction when the light enters the second plate ?
(b) When the light reaches the top of the second plate will it pass through the boundary? Justify your answer with a calculation.

Answers:

(a)

1 x sin 60 1.5 x sin r1 → r1 = 35.26^o

r1 = i2

1.5 sin i2 = 2.18 sin r2 → r2 = 23.4^o

r2 = 23.4^o

(b)

r2 = i3

2.18 sin i3 = 1 x sin r3 = sin r3

sin r3 ≤ 1 → 2.18 sin i3 ≤ 1/2.18 = 0.4587
→ sin i3 ≤ 27.30^o

But i3 = 66.6^o larger than the critical angle θc = 27.30^o

When the light reaches the top of the second plate, il will not pass through the boundary.

A 10.0 cm tall object is placed 80.0 cm in front of a concave mirror with a focal length of 30.0 cm.

(a) What are the image distance and the image type (real or virtual) ?
(b) What are the image height and the orientation of the image (upright or inverted) ?
(c) Sketch a ray-diagram for this situation directly on the figure above. Indicate where the image is formed.
(d) How will the image change as the object is slowly moved towards the mirror eventually reaching its vertex? Briey explain.

Answers:

See the figure.

Part II: Multiple Choice Questions

Mirrors summary