A big olive (m=0.50 kg) lies at the origin of an x y...

A big olive $(m=0.50 \mathrm{~kg})$ lies at the origin of an $x y$ coordinate system, and a big Brazil nut $(M=1.5 \mathrm{~kg})$ lies at the point $(1.0,2.0) \mathrm{m} .$ At $t=0$, a force $\vec{F}_{o}=(2.0 \hat{\mathrm{i}}+3.0 \mathrm{j}) \mathrm{N}$ begins to act on the olive, and a force $\vec{F}_{n}=(-3.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}) \mathrm{N}$ begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at $t=4.0 \mathrm{~s}$, with respect to its position at $t=0 ?$

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Key Concepts

Center of Mass

The center of mass is the weighted average position of a system's mass distribution. It represents the point at which the overall mass of the system can be thought of as being concentrated, and its motion is determined solely by the total external forces acting on the system, regardless of internal forces.

Newton’s Second Law for Systems

Newton’s second law extends to systems of particles by relating the net external force acting on the system to the acceleration of its center of mass. This law states that the sum of the external forces is equal to the total mass of the system multiplied by the acceleration of the center of mass.

Kinematics of Constant Acceleration

Under a constant net external force, the center of mass of a system experiences constant acceleration. The displacement of the center of mass can be found using kinematic equations, such as the equation that relates displacement to initial velocity, acceleration, and the square of the time elapsed.

Superposition of Forces

Superposition of forces is the principle that the net force acting on a system is the vector sum of all individual forces acting on the parts of the system. This net force determines the overall acceleration of the system's center of mass as calculated in Newton’s second law.

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