Physics Exam 1 Review

Displacement & Motion

100

This is the displacement when a runner completes a full lap around a circular track.

What is zero meters?

200

A jogger runs halfway around a circular track with radius 10 m. This is the magnitude of their displacement.

What is 20 meters (the diameter)?

300

You walk 0.8 mi east, then 0.7 mi at some angle south of east. If your total eastward displacement is 1.1 mi, this is the angle you walked relative to east on the second leg.

What is 64.6° south of east?

Eastward component: 0.7 cos(θ) = 1.1 - 0.8 = 0.3 mi

• cos(θ) = 0.3/0.7 = 0.429
• θ = 64.6°

 

400

This is the magnitude of your net displacement after walking 0.8 mi east, then 0.7 mi at 64.6° south of east.

What is 1.18 miles?

• Solution: Southward component: 0.7 sin(64.6°) = 0.632 mi
  • Net displacement = √(1.1² + 0.632²) = √(1.21 + 0.40) = 1.27 mi

500

After walking 0.8 mi east, then 0.7 mi at 64.6° south of east (net displacement 1.27 mi), this is the direction of your net displacement relative to east.

What is 29.8° south of east?

• Solution: tan(φ) = 0.632/1.1 = 0.575
  • φ = tan⁻¹(0.575) = 29.8°

Kinematics

100

On the Moon with gravity of 1.62 m/s², a rock dropped into a 120 m deep crater takes this many seconds to hit the bottom.

What is 12.2 seconds?

• Solution: Using d = ½gt²
  • 120 = ½(1.62)t²
  • t² = 148.1
  • t = 12.2 s

200

A car traveling at 30.0 m/s brakes and stops in 120 m. This was the car's acceleration.

What is -3.75 m/s²?

• Solution: Using v² = v₀² + 2ad
  • 0 = (30.0)² + 2a(120)
  • -900 = 240a
  • a = -3.75 m/s²

300

A car traveling at 30.0 m/s brakes with acceleration of -3.75 m/s² until it stops. This is how long it took.

What is 8.0 seconds?

• Solution: Using v = v₀ + at
  • 0 = 30.0 + (-3.75)t
  • t = 30.0/3.75 = 8.0 s

400

An astronaut drops a rock on the Moon (g = 1.62 m/s²) and it takes 12.2 seconds to hit the bottom. This is the speed of the rock just before impact.

What is 19.8 m/s?

• Solution: Using v = v₀ + gt
  • v = 0 + (1.62)(12.2)
  • v = 19.8 m/s

500

A car braking from 30.0 m/s stops in 120 m. At the halfway point (60 m), the car is traveling at this speed.

What is 21.2 m/s?

• Solution: Using v² = v₀² + 2ad with a = -3.75 m/s²
  • v² = (30.0)² + 2(-3.75)(60)
  • v² = 900 - 450 = 450
  • v = 21.2 m/s

FORCES IN ONE DIMENSION

100

A 300,000 kg aircraft traveling at 27.0 m/s experiences a braking force of 445,000 N. This is its acceleration.

 What is 1.48 m/s² (or -1.48 m/s² with direction)?

• Solution: a = F/m = 445,000/300,000 = 1.48 m/s²

200

 A 300,000 kg aircraft traveling at 27.0 m/s experiences a braking force of 445,000 N. This is its speed after 11.0 seconds.

What is 10.7 m/s?

• Solution:
  • a = -1.48 m/s²
  • v = v₀ + at = 27.0 + (-1.48)(11.0) = 10.7 m/s

300

A 10-kg object hangs from a wire in an upward-moving elevator. If the tension is 75 N, this is the magnitude of the elevator's acceleration.

What is 2.3 m/s²?

• Solution:
  • Weight: W = mg = 10(9.8) = 98 N
  • Net force: F_net = T - W = 75 - 98 = -23 N
  • a = F/m = -23/10 = -2.3 m/s²
  • Magnitude = 2.3 m/s²

400

A 10-kg object in an elevator has a tension of 75 N (less than its weight of 98 N). The elevator's acceleration is in this direction.

What is downward?

• Solution: Since T < W, the net force is downward, so acceleration is downward

500

A 300,000 kg aircraft braking at 445,000 N starts at 27.0 m/s. This is the total distance it travels before coming to a complete stop.

What is 247 meters?

• Solution:
  • a = -1.48 m/s²
  • Using v² = v₀² + 2ad with v = 0
  • 0 = (27.0)² + 2(-1.48)d
  • d = 729/2.96 = 246.3 m ≈ 247 m

FORCES IN TWO DIMENSIONS

100

Deetra (75 kg total) accelerates at 2 m/s² on frictionless ice. This thrust force is needed from her jet-powered skates.

What is 150 N?

• Solution: F = ma = 75(2) = 150 N

200

Deetra (75 kg) accelerates at 2 m/s² on ice with μ_k = 0.40. This is the friction force acting on her.

What is 294 N?

• Solution: f_k = μ_k mg = 0.40(75)(9.8) = 294 N

300

Deetra (75 kg) accelerates at 2 m/s² on ice with μ_k = 0.40. This thrust force is needed from her jet-powered skates.

What is 444 N?

400

A mass M hangs at rest from two wires meeting at a knot. One horizontal wire has tension 38 N, and an angled wire has tension 59 N. This is the angle the angled wire makes with the ceiling.  

What is 40.0°?

• Solution: At equilibrium, horizontal components balance:
  • 59 sin(θ) = 38
  • sin(θ) = 38/59 = 0.644
  • θ = 40.0°

 

500

 With a horizontal wire tension of 38 N and an angled wire (at 40.0° from ceiling) with tension 59 N, this is the mass of the hanging object.

What is 4.61 kg?

• Solution: Vertical equilibrium:
  • Weight = 59 cos(40.0°) = 59(0.766) = 45.2 N
  • Mass = 45.2/9.8 = 4.61 kg

CONCEPTUAL PHYSICS

100

This quantity is always measured as a straight line from starting position to ending position, regardless of the path taken.

What is displacement?

200

When an object in an elevator experiences less tension than its weight, the elevator must be accelerating in this direction.

What is downward?

300

On rough surfaces, to accelerate forward, the applied force must overcome this force plus provide the net force for acceleration.

What is the friction force (or kinetic friction)?

400

Unlike kinetic friction between solid surfaces, this type of resistive force does depend on the speed of the object.

What is air resistance (or drag)?

500

When a mass hangs at rest from multiple wires at different angles, this principle allows us to analyze the forces at the knot where the wires meet.

What is equilibrium (or the sum of forces equals zero)?