- A person walks forward inside a cart on a level, low-friction track. Consider the system (person+cart). What happens to the system's center-of-mass velocity?
Answer: It remains constant because forces between person and cart are internal to the system.
This problem demonstrates center-of-mass motion with internal forces in a horizontal system. The center-of-mass velocity of a system changes only when external forces act on it. As the person walks forward on the cart, they push the cart backward (Newton's third law), but these are internal forces within the person-cart system. On a level, low-friction track, there's negligible external horizontal force. Therefore, the center-of-mass velocity remains constant at whatever value it had initially. The person moves forward relative to the ground while the cart moves backward, but their combined center of mass maintains constant velocity. Choice D incorrectly assumes the center of mass must stop when objects move oppositely. To solve such problems, recognize that internal forces cause relative motion within the system but preserve center-of-mass motion.
- Two ice skaters stand on frictionless ice and push off each other. Define the system as both skaters together; the push forces they exert on each other are internal. A steady wind exerts a horizontal force to the east on both skaters (an external force on the system). Immediately after they push off, the skaters move apart while the wind continues. Which describes the motion of the system's center of mass?
Answer: It accelerates east because there is a net external force east.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Internal forces, such as the push between the skaters, cancel out and do not affect the center-of-mass motion. Here, the wind provides a net external force eastward on the system, causing the center of mass to accelerate east. Choice C is incorrect because while the internal pushes are equal and opposite, the external wind force still acts, leading to acceleration. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- A person stands on a skateboard on level ground. Define the system as person + skateboard. The person throws a heavy ball forward; the force between the person and the ball during the throw is internal only if the ball is included, but the ball is not in the system. Neglect air resistance, but assume the ground exerts negligible horizontal force on the skateboard. Initially the person and skateboard are at rest. After the ball leaves the person's hands, what happens to the center of mass of the defined system?
Answer: It moves backward because the ball exerts an external impulse on the system.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Since the ball is not part of the system, the force it exerts on the person during the throw is external, providing a backward impulse. With negligible other external horizontal forces, this causes the center of mass to move backward after the throw. Choice A is incorrect because the forces between person and skateboard are internal, but the external impulse from the ball causes motion. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- A magnet attracts a steel cart on a low-friction track. The magnet is mounted to a second cart; the carts pull toward each other and eventually collide. Define the system as both carts (including the magnet). The magnetic forces between the carts are internal. Assume external horizontal forces are negligible. Initially both carts are at rest. What is the correct description of the center-of-mass motion?
Answer: It remains at rest while the carts accelerate toward each other.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Internal magnetic forces between the carts cancel in pairs and do not produce net force on the system. With negligible external horizontal forces and initial rest, the center of mass remains at rest as the carts approach. Choice A is incorrect because magnetic forces are internal and cannot move the center of mass without external influence. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- A person stands on a skateboard and throws a ball forward; with person+skateboard+ball as the system, how does the center of mass behave while the ball is in flight?
Answer: It moves with constant horizontal velocity because external horizontal forces on the system are negligible.
This question tests understanding of center-of-mass motion for systems with only internal forces. When a person throws a ball while standing on a skateboard, the throwing force is internal to the person+skateboard+ball system. The center of mass of a system moves according to the net external force acting on it. Since there are negligible external horizontal forces (assuming a level surface with minimal friction/air resistance), the system's center of mass maintains constant horizontal velocity throughout the ball's flight. Choice A incorrectly suggests the ball can exert an external force on its own system after release, which is impossible—all ball-person interactions are internal. To analyze center-of-mass motion, first define your system boundaries, then identify which forces cross those boundaries (external) versus which act within them (internal).
- Two students on frictionless ice push off each other and then glide apart. Taking the system as both students together, the push forces are internal, while vertical normal forces and weights are external but cancel. Immediately after they push off, which statement about the center-of-mass motion is correct?
Answer: The center of mass remains at rest (or continues at constant velocity) because the net external force is zero.
This question tests understanding of center-of-mass motion for systems. The center of mass of a system accelerates only when there is a net external force acting on the system. In this scenario with two students pushing off each other on frictionless ice, the push forces between them are internal forces within the system of both students. The external forces (weights and normal forces) cancel vertically, and there are no external horizontal forces. Therefore, the center of mass remains at rest or continues with constant velocity, following Newton's first law. Choice A incorrectly suggests the center of mass favors the heavier student, but internal forces cannot change center-of-mass motion. To solve center-of-mass problems, always identify the system boundaries and check whether forces are internal or external to that system.
- A person walks from the stern to the bow of a floating boat. Define the system as person + boat; water exerts negligible horizontal force on the boat. As the person walks forward relative to the boat, the system's center of mass:
Answer: remains at rest (or constant velocity) relative to the shore because net external horizontal force is negligible.
This question tests understanding of center-of-mass motion for systems. The center of mass of a system accelerates only when there is a net external force acting on the system. In this boat scenario, the system is the person plus boat, and the water exerts negligible horizontal force. As the person walks forward relative to the boat, the boat moves backward relative to the shore due to momentum conservation. These are internal interactions that cannot change the center-of-mass position relative to the shore. Choice A incorrectly focuses only on the person's motion without considering the boat's compensating backward motion. To analyze relative motion within systems, remember that internal forces cause equal and opposite momentum changes that keep the center of mass fixed relative to external references.
- Two identical pucks glide on frictionless ice connected by a light spring; for the two-puck system, how does the center of mass behave as the spring oscillates?
Answer: It stays fixed or moves at constant velocity because the spring force is internal to the system.
This question tests understanding of center-of-mass motion for systems with only internal forces. When two pucks connected by a spring oscillate on frictionless ice, the spring force is internal to the two-puck system—it acts between system components. The center of mass of any system moves according to net external forces only, remaining unaffected by internal forces. Since the ice is frictionless and no external horizontal forces act on the system, the center of mass stays fixed (if initially at rest) or moves at constant velocity, even as the pucks oscillate dramatically. Choice A incorrectly suggests the alternating spring force moves the center of mass back and forth, but internal forces always sum to zero by Newton's third law. Remember: internal forces can create complex relative motions within a system while the center of mass follows simple motion determined only by external forces.
- A rocket in deep space (negligible external forces) fires an engine that expels exhaust backward. For the system (rocket+exhaust), what happens to the center-of-mass velocity?
Answer: It remains constant because all forces between rocket and exhaust are internal.
This question tests understanding of center-of-mass motion in isolated systems. The center of mass of an isolated system (one with negligible external forces) moves at constant velocity, as stated by Newton's first law applied to the system. In deep space, the rocket-exhaust system experiences negligible external forces. The engine creates forces between the rocket and exhaust, but these are internal to the system. While the rocket accelerates forward and exhaust accelerates backward, the center of mass of the combined system maintains its original velocity. Choice A incorrectly treats the engine thrust as external, but it's actually an internal force pair between rocket and exhaust. To analyze propulsion problems, always define your system to include all interacting parts, making thrust forces internal.
- Two masses are connected by a light rigid rod and slide on a frictionless horizontal table. The system boundary includes both masses and the rod. A student briefly applies a horizontal force to one mass, then removes it. During the push, the applied force is external to the system; forces in the rod are internal. After the student stops pushing, what happens to the center-of-mass motion of the system?
Answer: It continues moving at constant velocity in the direction it had at release.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. After the external push ends, internal forces in the rod cancel and do not change the system's momentum. With no ongoing external forces on the frictionless table, the center of mass continues at constant velocity. Choice C is incorrect because internal forces like rod tension cannot accelerate the center of mass. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- A firework rises vertically and explodes into two equal fragments that fly apart horizontally. Neglect air resistance. How does the center-of-mass motion change at the explosion?
Answer: It continues as if no explosion occurred, since the explosion forces are internal.
This question tests understanding of center-of-mass motion during internal explosions. The center of mass of a system follows a trajectory determined solely by external forces, regardless of internal interactions. Before the explosion, the firework's center of mass follows a parabolic path under gravity (neglecting air resistance). The explosion forces are internal to the firework system—they're forces between parts of the firework itself. These internal forces cause the fragments to fly apart but cannot change the center-of-mass trajectory. After explosion, the center of mass continues on the same parabolic path it would have followed if no explosion occurred. Choice A incorrectly assumes that opposite motions of fragments affect the center of mass. Remember: internal forces can rearrange mass within a system but never change the system's center-of-mass motion.
- Two ice skaters initially at rest push off each other on nearly frictionless ice. For the system of both skaters, what is true of the center of mass?
Answer: It remains at rest because there is no net external horizontal force.
This problem illustrates center-of-mass motion when only internal forces act. The center of mass of a system remains at rest (or moves at constant velocity) when the net external force is zero. The pushing forces between the skaters are internal to the two-skater system. Since the ice is nearly frictionless, there's no significant external horizontal force. The skaters will move apart with momenta that are equal and opposite, but their center of mass stays fixed at its initial position. Choice C incorrectly assumes the center of mass must be located inside one of the objects, but it's actually at a point between them. To analyze such problems, first identify your system boundary, then classify all forces as internal or external to that system.
- Two identical pucks on frictionless ice move toward each other with equal speed and stick together in a perfectly inelastic collision. The system boundary includes both pucks; contact forces during collision are internal, and external horizontal forces are negligible. After they stick, how does the center of mass move?
Answer: It remains at rest because the total momentum of the system is zero.
This question examines center-of-mass motion in collisions with no external forces. The motion of the center of mass of a system depends only on the net external force acting on the system. For the two pucks, there is no net external horizontal force on the frictionless ice, and initial total momentum is zero due to equal and opposite velocities. Thus, the center of mass remains at rest after they stick together in the inelastic collision. A common distractor is choice C, which mistakenly claims the center of mass accelerates during the collision due to large internal forces, but internal forces cannot change the center-of-mass velocity. To approach such problems, calculate initial total momentum and use conservation laws when external forces are negligible.
- A cart of mass M on a frictionless track carries a smaller block of mass m on top. The block is pushed by the cart's internal spring launcher so that the block slides to the right relative to the cart. Define the system as cart + block + launcher. All contact forces between cart and block are internal to the system; there are no external horizontal forces. What is the correct description of the center-of-mass motion?
Answer: It remains at rest because the net external horizontal force is zero.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Internal forces from the launcher and between cart and block cancel in pairs, producing no net force on the system. Without external horizontal forces, the center of mass remains at rest while the parts move oppositely. Choice A is incorrect because relative motion does not imply center-of-mass acceleration without external forces. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- Two blocks m1 and m2 rest on a rough horizontal floor and are connected by a light string. A student pulls on block m1 with a constant horizontal force to the right. Take the system boundary to include both blocks and the string. Friction from the floor on each block is external to the system. Which statement about the center-of-mass acceleration is correct?
Answer: It is determined by the net external horizontal force on the two-block system.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Internal forces, such as the tension in the string, do not affect the net force on the system. The net external horizontal force includes the pulling force and friction on both blocks, which determines the center-of-mass acceleration. Choice C is incorrect because internal forces like tension do cancel, but the acceleration is due to external forces, not zero unless those are balanced. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- A person walks from the stern to the bow of a floating boat; for the person+boat system, how does the center-of-mass motion change?
Answer: The center of mass stays at constant velocity (approximately zero) because horizontal external forces are negligible.
This question tests understanding of center-of-mass motion when objects rearrange within a system. When a person walks from stern to bow on a floating boat, the forces between person and boat are internal to the person+boat system. The center of mass of a system only accelerates if there's a net external force, and since water resistance is negligible (implied by "floating"), the horizontal external forces are approximately zero. Therefore, the center of mass maintains constant velocity, which is approximately zero if the system started at rest. As the person moves forward relative to the water, the boat moves backward to keep the center of mass stationary. Choice B incorrectly assumes the center of mass follows the person's motion, ignoring that the boat must move oppositely. To analyze rearrangement problems, focus on whether external forces exist rather than tracking individual object motions.
- A ball and Earth interact gravitationally as the ball falls straight down. Define the system as ball + Earth; the gravitational forces between them are internal, and ignore air resistance and other external forces. What happens to the center-of-mass motion of this system?
Answer: It remains at rest (or constant velocity) because there is no net external force on the system.
This question tests understanding of center-of-mass motion for systems. The center of mass of a system accelerates only when there is a net external force acting on the system. With the system defined as ball plus Earth, the gravitational forces between them are internal forces—the ball pulls Earth up while Earth pulls the ball down with equal magnitude. Since these are the only significant forces and they are internal to the system, there is no net external force. Therefore, the center of mass of the ball-Earth system remains at rest or continues at constant velocity. Choice A incorrectly considers only the ball's motion without recognizing that Earth also accelerates (imperceptibly) toward the ball. To properly analyze gravitational systems, include all interacting objects and recognize that their mutual forces are internal to the complete system.
- Two identical carts move toward each other on a frictionless track with equal speeds. They collide and stick. For the two-cart system, the center-of-mass velocity after collision is
Answer: zero because the net external horizontal force is zero and initial total momentum is zero.
This problem demonstrates center-of-mass motion in symmetric collisions. The center-of-mass velocity equals the total momentum divided by total mass. Initially, the two identical carts have equal speeds in opposite directions, giving them equal but opposite momenta. The total initial momentum is therefore zero. During the collision, the forces between carts are internal to the two-cart system. On a frictionless track, there's no external horizontal force to change the total momentum. Since momentum remains zero and the carts stick together, they must be at rest after collision—meaning the center-of-mass velocity is zero. Choice A incorrectly relates energy loss to center-of-mass motion; these are independent concepts. When analyzing collisions, always check if the initial momentum is zero—if so, the center of mass remains stationary regardless of collision type.
- Two identical pucks slide on frictionless ice. They collide and stick together. Choose the system as both pucks together; the contact forces during the collision are internal. There are no external horizontal forces. Before the collision, puck 1 moves east and puck 2 moves west with equal speed. Which statement about the center-of-mass motion is correct?
Answer: The center of mass remains at rest before, during, and after the collision.
This question tests the concept of center-of-mass motion for a system of particles. The motion of the center of mass of a system is determined solely by the net external force acting on the system, as if all the mass were concentrated at the center of mass and all external forces acted there. Internal forces during the collision cancel out and do not affect the center-of-mass velocity. With no external horizontal forces and initial total momentum zero, the center of mass remains at rest throughout. Choice D is incorrect because internal forces cannot create net momentum; the stuck pucks stop, but the center of mass stays put. To solve similar problems, always identify the system boundary and calculate the net external force to determine the center-of-mass acceleration.
- Two students on carts (A heavier than B) start at rest on frictionless level track and push each other apart. The system boundary includes both carts and students; the push forces are internal, and vertical normal and weight forces are external but cancel. After they separate, which statement about the system's center-of-mass motion is correct?
Answer: The center of mass remains at rest because there is no net external horizontal force on the system.
This question assesses understanding of center-of-mass motion in isolated systems. The motion of the center of mass of a system depends only on the net external force acting on the system. In this scenario, the system of both carts and students experiences no net external horizontal force, as the track is frictionless and vertical forces cancel. Therefore, the center of mass, which was initially at rest, must remain at rest even after the students push apart, as internal push forces do not affect the overall center-of-mass motion. A common distractor is choice A, which incorrectly suggests that internal forces can cause the center of mass to accelerate toward the heavier cart, but internal forces cancel out in pairs and do not influence the center-of-mass acceleration. To analyze similar problems, always identify the system boundary and determine the net external force to predict center-of-mass behavior.
