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Inertia mass and density

Learning objectives

  • Outcome

    Calculate the acceleration of a body from the net force and its inertial mass using a=F/ma=F/m, stating the units.

  • Outcome

    Calculate density from mass and volume using ρ=m/V\rho=m/V and mass from density and volume using m=ρVm=\rho V.

  • Outcome

    Compare the motions of two bodies with equal volumes but different densities under the same net force, and justify which one changes speed more slowly.

Understand the idea

Inertial Mass and Density: How Material Properties Determine Motion

Text

Speed and average speed describe how position changes over time, but they do not explain why the same push changes the speed of one body more than another. The inertial mass mm of a body measures its resistance to a change in velocity: under the same net force FF, a larger mass has a smaller acceleration, a=F/ma=F/m. Density ρ\rho connects inertial mass to volume for a uniform substance: ρ=m/V\rho=m/V, so m=ρVm=\rho V and V=m/ρV=m/\rho. Thus mass and density are different properties: two bodies can have the same mass but different densities if their volumes differ, and the same volume but different masses if their densities differ. In calculations, keep units consistent; in SI, mm is in kilograms, VV in cubic metres, ρ\rho in kilograms per cubic metre, FF in newtons, and aa in metres per second squared. For constant net force, a denser body of the same volume has more inertial mass and therefore changes its speed more slowly.

Worked examples

Direct

Problem

A cart has inertial mass m=2.0 kgm=2.0\,\mathrm{kg}. A net force F=6.0 NF=6.0\,\mathrm{N} acts on it along the direction of motion. Calculate its acceleration and state the formula used.

Solution

Use Newton's second law in the form a=F/ma=F/m: a=Fm=6.0 N2.0 kg=3.0 m/s2.a=\frac{F}{m}=\frac{6.0\,\mathrm{N}}{2.0\,\mathrm{kg}}=3.0\,\mathrm{m/s^2}. The acceleration is in the same direction as the net force.

Application

Problem

A solid cube has volume V=1.0×10−3 m3V=1.0\times10^{-3}\,\mathrm{m^3} and density ρ=2500 kg/m3\rho=2500\,\mathrm{kg/m^3}. A net force F=5.0 NF=5.0\,\mathrm{N} acts on it. Calculate its inertial mass and acceleration.

Solution

First find the mass from density: m=ρV=2500 kg/m3×1.0×10−3 m3=2.5 kg.m=\rho V=2500\,\mathrm{kg/m^3}\times1.0\times10^{-3}\,\mathrm{m^3}=2.5\,\mathrm{kg}. Then find the acceleration: a=Fm=5.0 N2.5 kg=2.0 m/s2.a=\frac{F}{m}=\frac{5.0\,\mathrm{N}}{2.5\,\mathrm{kg}}=2.0\,\mathrm{m/s^2}.

Transfer

Problem

Two spheres have the same volume V=5.0×10−4 m3V=5.0\times10^{-4}\,\mathrm{m^3}. One is aluminium with ρA=2700 kg/m3\rho_A=2700\,\mathrm{kg/m^3}; the other is plastic with ρP=900 kg/m3\rho_P=900\,\mathrm{kg/m^3}. Both move at initial speed v0=2.0 m/sv_0=2.0\,\mathrm{m/s} and are slowed by the same constant resistive force F=0.90 NF=0.90\,\mathrm{N}. Assuming only this force acts, determine which sphere travels farther before stopping and calculate the stopping distance for each. Use d=v02/(2a)d=v_0^2/(2a).

Solution

Find the masses: mA=ρAV=2700×5.0×10−4=1.35 kg,m_A=\rho_A V=2700\times5.0\times10^{-4}=1.35\,\mathrm{kg}, mP=ρPV=900×5.0×10−4=0.45 kg.m_P=\rho_P V=900\times5.0\times10^{-4}=0.45\,\mathrm{kg}. With the same resistive force, the decelerations are aA=FmA=0.901.35=23 m/s2,a_A=\frac{F}{m_A}=\frac{0.90}{1.35}=\frac{2}{3}\,\mathrm{m/s^2}, aP=FmP=0.900.45=2.0 m/s2.a_P=\frac{F}{m_P}=\frac{0.90}{0.45}=2.0\,\mathrm{m/s^2}. The stopping distances are dA=v022aA=(2.0)22(2/3)=3.0 m,d_A=\frac{v_0^2}{2a_A}=\frac{(2.0)^2}{2(2/3)}=3.0\,\mathrm{m}, dP=v022aP=(2.0)22(2.0)=1.0 m.d_P=\frac{v_0^2}{2a_P}=\frac{(2.0)^2}{2(2.0)}=1.0\,\mathrm{m}. The aluminium sphere travels farther. For the same volume it is denser, so it has more inertial mass; the same force gives it a smaller deceleration.

Practice

Text

Use a safe interactive simulation such as PhET Forces and Motion: Basics or Density.

  1. In the force-and-motion mode, set a constant applied force, record the acceleration for carts of mass 1 kg1\,\mathrm{kg}, 2 kg2\,\mathrm{kg}, and 4 kg4\,\mathrm{kg}; plot or tabulate acceleration versus mass.
  2. In the density mode, choose a material and a fixed volume, read the mass, and verify m=ρVm=\rho V.
  3. Repeat for two materials with the same volume but different densities; predict which body will stop later under the same resistive force, then test the prediction in the motion mode.
  4. Write a table of mm, VV, ρ\rho, FF, and aa, and check that all units are SI.

Visual reference

Purpose

To show that inertial mass controls acceleration under a given force, while density connects mass to volume for a material.

Representation

A two-panel diagram. Left: two identical carts, one with mass mm and one with mass 2m2m, each pulled by the same force FF; acceleration arrows show a=F/ma=F/m and a/2a/2. Right: two cubes of equal volume VV, one with low density ρ1\rho_1 and one with high density ρ2>ρ1\rho_2>\rho_1; labels show m1=ρ1Vm_1=\rho_1 V and m2=ρ2Vm_2=\rho_2 V, and the cubes have the same size but different mass markers.

Real-world examples

Football and shot put

Scenario

A football has mass m1=0.43 kgm_1=0.43\,\mathrm{kg} and flies at average speed v1=25 m/sv_1=25\,\mathrm{m/s}. A shot put used in athletics has mass m2=7.26 kgm_2=7.26\,\mathrm{kg} and leaves the hand at v2=13 m/sv_2=13\,\mathrm{m/s}. A goalkeeper catching the football feels a small change of motion; a field judge stopping the shot put must exert a much larger force over a longer time. During uniform slowing over 1.0 s1.0\,\mathrm{s}, the football's average speed changes by 25 m/s25\,\mathrm{m/s}, while the shot's changes by 13 m/s13\,\mathrm{m/s}.

Connection

The scenario exercises inertial mass: for the same kind of velocity change, the 7.26 kg7.26\,\mathrm{kg} shot resists a change in its state of motion more than the 0.43 kg0.43\,\mathrm{kg} football. The average speed relation vavg=Δs/Δtv_{\mathrm{avg}}=\Delta s/\Delta t tells how quickly each body moves before the catch, so the comparison is about mass, not just speed.

Truck with sand

Scenario

An empty truck has mass 12000 kg12000\,\mathrm{kg} and takes 20 m320\,\mathrm{m^3} of sand with density 1600 kg/m31600\,\mathrm{kg/m^3}. The sand mass is m=ρV=1600⋅20=32000 kgm=\rho V=1600\cdot20=32000\,\mathrm{kg}, so the total mass of the loaded truck is 12000+32000=44000 kg12000+32000=44000\,\mathrm{kg}. Both the empty and the loaded truck travel at v=15 m/sv=15\,\mathrm{m/s}. During a uniform stop in 5.0 s5.0\,\mathrm{s}, the loaded truck requires a much larger braking force; the empty truck changes speed more easily.

Connection

Here density ρ=m/V\rho=m/V converts the sand volume into mass, and mass acts as the measure of inertia: the loaded truck of mass 44000 kg44000\,\mathrm{kg} resists a change in speed more than the empty one of mass 12000 kg12000\,\mathrm{kg}. Average speed vavg=s/tv_{\mathrm{avg}}=s/t gives the same initial motion, so the difference during braking is set by mass.

Oil and water in a bottle

Scenario

A bottle of volume 1.0 L=1.0×10−3 m31.0\,\mathrm{L}=1.0\times10^{-3}\,\mathrm{m^3} is filled with water of density 1000 kg/m31000\,\mathrm{kg/m^3}; the water mass is 1000⋅1.0×10−3=1.0 kg1000\cdot1.0\times10^{-3}=1.0\,\mathrm{kg}. The same bottle is filled with sunflower oil of density 920 kg/m3920\,\mathrm{kg/m^3}; the oil mass is 920⋅1.0×10−3=0.92 kg920\cdot1.0\times10^{-3}=0.92\,\mathrm{kg}. If both bottles are pushed across a table at average speed 0.50 m/s0.50\,\mathrm{m/s} and then stopped, the water bottle resists the change in motion slightly more.

Connection

The scenario links density and inertial mass: for the same volume 1.0 L1.0\,\mathrm{L}, density ρ=m/V\rho=m/V gives different masses 1.0 kg1.0\,\mathrm{kg} and 0.92 kg0.92\,\mathrm{kg}. The average speed before stopping is the same, so inertial mass determines which bottle is harder to change in speed.