- Basic Math and Science Newton's Laws of Motion Questions and Answers Flashcards

A 2 kg block rests on a frictionless surface and is connected via a massless string over a frictionless pulley to a 0.5 kg hanging mass. What is the acceleration of the system when released?

Answer: 1.96 m/s²

Using Newton's second law for the system: net force = 0.5 × 9.8 = 4.9 N acts on a total mass of 2 + 0.5 = 2.5 kg. Acceleration = 4.9 / 2.5 = 1.96 m/s². The entire system (both masses) accelerates together, so the total inertia resisting the gravitational pull on the hanging mass must be accounted for.

According to Newton's Third Law, when a horse pulls a cart forward, the cart pulls the horse backward with an equal and opposite force. Why does the horse-cart system still accelerate forward?

Answer: The action-reaction forces act on different objects, so the net force on the system is the ground's friction pushing the horse forward

Action-reaction force pairs always act on DIFFERENT objects, so they never cancel each other for the purpose of calculating net force on a single object or system. The horse pushes backward on the ground; the ground pushes forward on the horse (reaction). This ground-reaction force on the horse exceeds the road friction on the cart, producing a net forward force on the horse-cart system.

A rocket in deep space (no gravity, no air resistance) is moving at constant velocity. The engines fire and exert a constant thrust force for 10 seconds, then shut off. Which of the following best describes the rocket's motion immediately after the engines shut off?

Answer: The rocket continues at the new, higher constant velocity

Newton's First Law states that an object in motion remains in motion at constant velocity unless acted upon by a net external force. In deep space with no gravity or air resistance, once the engines shut off there is zero net force. The rocket retains the velocity it reached at the end of the thrust phase and maintains it indefinitely.

Two ice skaters push off each other from rest. Skater A has a mass of 80 kg and moves away at 3 m/s. Skater B has a mass of 60 kg. What is Skater B's speed after the push?

Answer: 4 m/s

By conservation of momentum (derived from Newton's Third Law), total momentum before and after must be equal. Before: p = 0. After: (80)(3) + (60)(v_B) = 0 → 240 + 60v_B = 0 → v_B = −4 m/s. The magnitude of Skater B's speed is 4 m/s (in the opposite direction). The lighter skater moves faster to conserve momentum.

A 5 kg box is pushed across a floor with a horizontal applied force of 30 N. The coefficient of kinetic friction is 0.4. What is the net acceleration of the box? (g = 9.8 m/s²)

Answer: 2.08 m/s²

Normal force N = mg = 5 × 9.8 = 49 N. Kinetic friction force = μ_k × N = 0.4 × 49 = 19.6 N. Net force = Applied − Friction = 30 − 19.6 = 10.4 N. Acceleration = F_net / m = 10.4 / 5 = 2.08 m/s². Newton's Second Law requires subtracting the opposing friction force before dividing by mass.

An astronaut in a sealed, windowless spacecraft cannot feel any forces and observes objects floating freely beside her. Which combination of conditions could explain this observation according to Newtonian mechanics?

Answer: The spacecraft is either in free fall (orbiting) or moving at constant velocity far from any gravitational field

Newton's First Law tells us objects with no net force move at constant velocity (including floating in place relative to surroundings). This occurs in two Newtonian scenarios: (1) constant velocity far from gravity sources, or (2) free fall/orbit where gravitational force accelerates all objects — including the astronaut and her surroundings — equally, so there is no relative force felt. Both conditions produce the observed floating. A spacecraft accelerating upward would press objects toward the floor, not float them.