Physics Mid-Term Exam Revision: It Is Important ✓ Solved

Physics Mid-Term Exam Revision81914directionsit Is Important

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Introduction

The following paper provides a comprehensive exploration of fundamental concepts in physics as prompted by a midterm revision assignment. The questions span a wide range of topics, including Newtonian mechanics, gravitational forces, projectile motion, and basic experimental design, demanding both theoretical knowledge and practical understanding. This paper will address each question systematically, providing clear explanations, mathematical calculations where appropriate, and thoughtful analysis, synthesized from credible physics literature and principles.

Question 1: Misconception of 'theory' in scientific context

In a conversation about the Moon landing, Mia claims to have a "theory" on the matter. Scientifically, a theory is a well-substantiated explanation of some aspect of the natural world, based on a body of evidence and repeatedly tested hypotheses. Mia's use of "theory" implies a hypothesis or an unproven idea, which is incorrect in this context because the Moon landings are supported by extensive empirical evidence, making it a scientific fact, not merely a theory. Therefore, Mia's use of the term "theory" misunderstands its scientific meaning, suggesting speculation rather than established knowledge.

Question 2: Designing an experiment to estimate average speed

Choosing scenario (e), the tip of a swinging baseball bat, I would design an experiment using a high-speed camera and a marked grid. The baseball bat would be swung in a controlled environment, and the camera would record the motion from a side view. By analyzing the video frames, I could measure the distance traveled by the tip per unit of time. The materials needed would include a high-speed video camera, a measuring tape, and markers on the bat. The procedure involves recording multiple swings, measuring the displacement of the bat's tip over a known time interval, and calculating the average speed as displacement divided by time.

Question 3: Example of balanced forces with constant velocity

An example is a fish swimming at a constant speed in calm water. The forces acting on the fish include the thrust from its fins and tail, and the water's resistance. When these forces are balanced, the fish maintains a constant velocity. If the fish accelerates, it would increase its thrust, which would disrupt the balance and result in acceleration until the forces are again balanced. Conversely, increasing water resistance (e.g., in turbulent water) would slow the fish down if it doesn't increase thrust, illustrating how forces influence motion.

Question 4: Weight on Jupiter

The weight of a person is calculated by multiplying mass by gravitational acceleration. Given: weight on Earth (Wₑ) = 500 N, acceleration due to gravity on Earth (gₑ) = 9.8 m/s². The person's mass (m) is Wₑ / gₑ = 500 / 9.8 ≈ 51.02 kg. On Jupiter, gravity (gⱼ) = 26 m/s². Therefore, weight on Jupiter (Wⱼ) = m × gⱼ ≈ 51.02 × 26 ≈ 1325.5 N.

Question 5: Speed increase of a falling rock on a different planet

The acceleration due to gravity (g) = 20 m/s² means the velocity of a freely falling object increases by g m/s each second. Starting from rest, the velocity after 1 second is 20 m/s, after 2 seconds is 40 m/s, and so forth. Therefore, the speed readings would increase by 20 m/s every second, with the velocity after t seconds being v = g × t = 20 × t m/s.

Question 6: Jumping into the pool from a balcony

The calculation involves projectile motion principles. The time to fall from a height (h = 45 m) is t = √(2h / g) ≈ √(2×45 / 9.8) ≈ 3.03 seconds. To reach horizontally over 15 meters in this time, the necessary horizontal velocity is v = distance / time = 15 / 3.03 ≈ 4.95 m/s. Practically, jumping with this speed is challenging and potentially dangerous, as it requires precise timing, strong jumping velocity, and control to land safely in the pool.

Question 7: Measuring bullets fired from an aircraft

a. When fired directly ahead while the plane is moving, the bullet's eastward speed relative to an observer on the ground is 600 + 600 = 1200 mph.

b. Firing in the opposite direction (westward), the bullet's speed relative to ground is 600 - 600 = 0 mph; it appears to drop vertically.

c. Firing vertically downward, the bullet's horizontal speed remains 600 mph eastward, but combined with any vertical components, it moves east at 600 mph as it falls. These measurements are possible with appropriate instruments, considering the frame of reference.

Question 8: Drag force on parachuting skydiver

Given: mass (m) = 50 kg, upward acceleration (a) = 6.2 m/s², gravity (g) = 9.8 m/s². The net force (F_net) = m × a = 50 × 6.2 = 310 N. The weight (W) = m × g = 50 × 9.8 = 490 N. The drag force (F_drag) balances the excess of weight over the net force: F_drag = W - F_net = 490 - 310 = 180 N.

Question 9: Comparing forces in towing scenarios

In Scenario A, since the truck's mass is larger than the trailer's, the force exerted by the truck on the trailer is greater than the force exerted by the trailer on the truck during acceleration. In Scenario B, both masses are equal and moving at the same speed; forces are equal in magnitude. In Scenario C, the trailer's mass exceeds the truck's, but with a slower velocity, the truck's force on the trailer may slightly differ depending on the acceleration, but overall, the forces are comparable; more detailed calculations are needed for precise comparison.

Question 10: Impact forces and momentum change

a. True: Newton's third law states equal and opposite forces, so impact forces are equal in magnitude.

b. True: Impulse equals change in momentum; since the forces are equal and act over the same contact time, the impulses are equal.

c. False: The change in speed depends on mass; the bug's mass is tiny compared to the car, so their speed changes differ significantly.

d. True: The change in momentum (impulse) is equal and opposite for both.

e. False: The accelerations are not the same; they depend on mass, so the bug experiences a much higher acceleration.

Question 11: Force, work, and power during stair climbing

Betty and Bianca weigh the same, so their forces are identical. Betty climbs faster (30s vs. 40s), performing the same work but in less time—thus, her power output is greater. Work depends on the change in vertical height and weight, which are equal for both. Therefore, Betty has higher power due to her faster ascent even though forces and work are the same, as power is work divided by time.

Question 12: Calculating linear speed from wheel rotation

Given: circumference = 2 m, rotational speed = 1 revolution/sec. The linear speed v = circumference × rotations/sec = 2 m × 1 = 2 m/sec.

Question 13: Physics of a trapeze scene

The trapeze artist's motion involves inertia, as the body resists changes in motion. The centripetal acceleration acts toward the center of the circular path when the artist swings, caused by the tension in the ropes providing a force directed inward. Gravity acts downward, influencing the overall trajectory, while velocity determines the speed of swinging. These factors combine to produce the elegant, circular motion seen in trapeze acts.

Question 14: Height to weigh 1/16th of current weight

Weighing less corresponds to a reduction in gravitational acceleration: W = m × g, so W_new = W / 16 fits g_new = g × (r / R)^2, where R is Earth's radius. Solving for height h above Earth's surface involves the relation g_new = g (R / (R + h))^2. Setting g_new = g / 16 leads to (R / (R + h))^2 = 1/16. Taking square roots: R / (R + h) = 1/4. Therefore, R + h = 4R, so h = 3R. With R = 6400 km, h ≈ 3 × 6400 km = 19200 km.

Question 15: Gravitational force comparison between sun and moon

The gravitational force F = G × (M × m) / r². The sun's mass is 2.7 × 10^7 times Earth's mass, and the sun's distance is 400 times the moon's distance. Since force depends on M / r², the ratio of the forces is (2.7 × 10^7) / (400)² ≈ (2.7 × 10^7) / (160,000) ≈ 168.75. Hence, the sun exerts approximately 169 times greater gravitational force on Earth than the moon does.

Question 16: Kinetic energy at a different orbital point

Given: Potential energy PE₁=7000 MJ, KE₁=4000 MJ; PE₂=2000 MJ. Total energy at the first point is TE₁ = PE₁ + KE₁ = 11,000 MJ. Negotiating conservation of energy, total energy at the second point (TE₂) = PE₂ + KE₂. Assuming total energy remains constant, TE₂ = TE₁ = 11,000 MJ. Therefore, KE₂ = TE₂ - PE₂ = 11,000 - 2000 = 9,000 MJ.

Question 17: Inelastic collision velocity calculation

Initial: m₁ = unknown, m₂ = twice m₁, v₂ = 5 m/s, m₁ is stationary. Conservation of momentum: m₁ × 0 + 2m₁ × 5 = (m₁ + 2m₁) × v_f. Simplify: 10m₁ = 3m₁ × v_f. Then v_f = 10m₁ / 3m₁ = 10/3 ≈ 3.33 m/s.

Question 18: Average impact force on a bullet

Mass m = 2.0 g = 0.002 kg, initial velocity v = 3.00 × 10^4 cm/s = 300 m/s, penetration depth d = 0.05 m. Using work-energy principle: work done = initial kinetic energy = the average force × distance. Therefore: F_avg = (1/2) m v² / d = (0.5)(0.002)(300)² / 0.05 ≈ (0.001)(90000) / 0.05 ≈ 90 / 0.05 = 1800 N.

Question 19: Traveling at relativistic speeds and aging

According to Einstein's theory of special relativity, as an object moves near the speed of light, time dilation occurs, meaning that from the perspective of the traveler, less time passes—they age slower relative to an outside observer. Thus, theoretically, traveling at such speeds could result in the traveler aging less than those remaining stationary, effectively making them "younger" in comparison. However, achieving such speeds with human technology remains far beyond current capabilities, and immense energy is required, making practical travel at near-light speeds currently impossible.

Question 20: Path difference between light and baseball under gravity

Gravity bends the path of the baseball significantly because it has mass and is affected by gravitational force, resulting in a curved trajectory. Light, being massless, follows straight lines in space, but its path appears bent when passing near massive objects due to gravitational lensing, a result of spacetime curvature. Therefore, although both are influenced by gravity, the baseball's trajectory is primarily a classical curved path, while light's path is affected by spacetime geometry, leading to different observable effects.

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