Mechanics 8.3 - Weight and motion under gravity

Mechanics 8.3 - Weight and motion under gravity

Use weight and gravitational acceleration in straight-line motion. You will distinguish kilograms from newtons, choose signs for vertical motion, find greatest heights and impact speeds, and judge the accuracy and limits of a constant-gravity model.

Weight as a force

If one freely falling ball has twice the mass of another, must it accelerate twice as fast? A 2 kg2\,\mathrm{kg} ball has weight 19.6 N19.6\,\mathrm N, while a 4 kg4\,\mathrm{kg} ball has weight 39.2 N39.2\,\mathrm N. The force doubles, but so does the mass that must be accelerated. We will find what that means for their motion.

You already know

Newton's second law is ∑F=ma\sum\mathbf F=m\mathbf a. Mass is measured in kg\mathrm{kg}, force in N\mathrm N, and acceleration in m s−2\mathrm{m\,s^{-2}}.

Keep the mass and the force due to gravity separate, even when everyday speech calls both “weight”.

Mass and weight

Mass is the quantity that relates a body's resultant force to its acceleration. Weight is the gravitational force on the body, vertically downwards near Earth's surface. For example, 3.6 kg3.6\,\mathrm{kg} is a mass; 35.3 N35.3\,\mathrm N is a weight, not a mass.

The magnitude of gravitational acceleration is written gg. Its value is a positive magnitude; the sign of its component depends on our axis.

Weight

For mass mm in a region where gravitational acceleration has magnitude gg,

W=mg.W=mg.

Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} in Edexcel Mathematics unless another value is stated. Not in the formula booklet: learn it.

A force arrow labelled mgmg is not an acceleration arrow labelled gg.

[DIAGRAM: asset_slug: p3_8_3_understand_and_use_weight_and_motion_in_a_straight_line_under_gravity__diagram_01; asset_name: Mass weight and free-fall acceleration; recommended_method: drawn_math; caption: Weight is a force on a mass; free-fall acceleration is force divided by mass; description: Two separated panels. The left panel shows a particle of mass m kg with one downward force arrow W equals mg in newtons. The right panel shows downward free-fall acceleration g in metres per second squared and the equation a equals mg divided by m equals g for positive mass. Text states that only gravity acts in free fall and that g depends on location but is treated as constant in this model.]
Weight is a force on a mass

Use the default value of gg in both examples.

Worked example 1: mass to weight

Question. A parcel has mass 3.6 kg3.6\,\mathrm{kg}. Find its weight.

Plan. Multiply the mass by gg, then round the force at the end.

W=mgweight is gravitational force=3.6(9.8)=35.28 Nsubstitute in SI units\begin{aligned} W&=mg &&\text{weight is gravitational force}\\ &=3.6(9.8)=35.28\,\mathrm N &&\text{substitute in SI units} \end{aligned}

Why multiply by gg? Kilograms describe mass; newtons describe the resulting gravitational force.

Answer. W=35.3 NW=35.3\,\mathrm N, to 33 significant figures, vertically downwards.

Check. Since gg is close to 1010, the weight should be close to 36 N36\,\mathrm N.

Change the information given: now the weight is known and the mass is required. The relationship stays the same.

Worked example 2: weight to mass

Question. A particle has weight 24.5 N24.5\,\mathrm N. Find its mass.

Plan. Rearrange W=mgW=mg before substituting.

m=Wgdivide by gravitational acceleration=24.59.8=2.50 kgsubstitute\begin{aligned} m&=\frac Wg &&\text{divide by gravitational acceleration}\\ &=\frac{24.5}{9.8}=2.50\,\mathrm{kg} &&\text{substitute} \end{aligned}

Why divide? Weight already contains the factor gg; multiplying would apply gravity twice.

Answer. The mass is 2.50 kg2.50\,\mathrm{kg}.

Check. 2.5(9.8)=24.5 N2.5(9.8)=24.5\,\mathrm N.

Mass and weight also respond differently to a change of location: the same body's mass stays the same while its weight changes if gg changes.

Watch out

A 0.75 kg0.75\,\mathrm{kg} mass does not have weight 0.75 N0.75\,\mathrm N. Use W=mgW=mg; keep the units attached to the quantity you have calculated.

Find the weight of this new mass.

Gravity and signs

An upward-thrown ball has downward acceleration while travelling both up and down. At its highest point its velocity is zero for an instant, but gravity has not switched off.

[DIAGRAM: asset_slug: p3_8_3_understand_and_use_weight_and_motion_in_a_straight_line_under_gravity__diagram_02; asset_name: Mechanics 8.3 - Weight and motion under gravity - diagram 02; recommended_method: drawn_math; description: Draw a vertical straight-line motion diagram on a 16:9 white canvas. Show an upward positive axis from launch point O, an object projected upward with initial velocity u, a highest point T where v = 0, and a downward acceleration arrow labelled a = -g when upward is positive. Include a lower ground/impact level with a note that velocity just before impact is not generally zero. Use spacious labels and no clutter.]
Diagram

Choose the axis once for the whole flight, rather than reversing it when the ball starts falling.

Signed gravitational acceleration

For free vertical motion with constant downward gravitational acceleration,

a=−gif upward is positivea=+gif downward is positive\begin{aligned} a&=-g &&\text{if upward is positive}\\ a&=+g &&\text{if downward is positive} \end{aligned}

The magnitude gg stays positive. These sign choices are not printed as a rule in the booklet. Not in the formula booklet: learn it.

With upward positive, use negative velocity for downward motion. The constant-acceleration equation v=u+atv=u+at is in the formula booklet.

Worked example 3: while the ball is rising

Question. A ball is projected vertically upwards at 14 m s−114\,\mathrm{m\,s^{-1}}. Find its velocity after 0.60 s0.60\,\mathrm s, using g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Plan. Take upward positive, so u=14u=14 and a=−9.8a=-9.8.

v=u+atconstant acceleration=14−9.8(0.60)=8.12 m s−1substitute with signs\begin{aligned} v&=u+at &&\text{constant acceleration}\\ &=14-9.8(0.60)=8.12\,\mathrm{m\,s^{-1}} &&\text{substitute with signs} \end{aligned}

Why use a negative acceleration? Gravity points down while our axis points up.

Answer. 8.12 m s−18.12\,\mathrm{m\,s^{-1}} upwards.

Check. The positive velocity is smaller than the initial 1414, so the rising ball is slowing.

Change only the observation time. Keep the same axis even after the highest point.

Worked example 4: after the ball turns

Question. Find the velocity of the same ball after 2.0 s2.0\,\mathrm s.

Plan. Use the same initial velocity and acceleration at the later time.

v=14−9.8(2.0)same model and axis=−5.6 m s−1evaluate\begin{aligned} v&=14-9.8(2.0) &&\text{same model and axis}\\ &=-5.6\,\mathrm{m\,s^{-1}} &&\text{evaluate} \end{aligned}

Why retain the minus sign? It identifies downward velocity; the speed is the magnitude.

Answer. The velocity is −5.6 m s−1-5.6\,\mathrm{m\,s^{-1}} with upward positive, or 5.6 m s−15.6\,\mathrm{m\,s^{-1}} downwards.

Check. The turning time is 14/9.8=1.428… s14/9.8=1.428\ldots\,\mathrm s, earlier than 2.0 s2.0\,\mathrm s, so downward motion is consistent.

A dropped particle follows the same sign rule: its zero initial velocity does not make its acceleration zero.

Watch out

“At the highest point, a=0a=0” is wrong. The velocity is zero there; the acceleration is still −g-g with upward positive.

Interpret the sign in another upward launch.

Free vertical motion with SUVAT

In free motion under gravity, ignore air resistance and treat the particle as moving with constant acceleration downwards. A particle launched upwards at 12 m s−112\,\mathrm{m\,s^{-1}} reaches a point where its vertical velocity is zero. That gives the missing information needed to calculate its greatest height.

Constant-acceleration equations

For motion along one fixed axis with constant signed acceleration,

v=u+ats=ut+12at2s=vt−12at2v2=u2+2ass=12(u+v)t\begin{aligned} v&=u+at\\ s&=ut+\tfrac12at^2\\ s&=vt-\tfrac12at^2\\ v^2&=u^2+2as\\ s&=\tfrac12(u+v)t \end{aligned}

Here ss is displacement from the initial point, not total distance. In the formula booklet.

Choose an equation that omits the quantity you do not need. For greatest height without a time, v2=u2+2asv^2=u^2+2as is direct.

Worked example 5: greatest height

Question. A particle is projected vertically upwards from OO at 12 m s−112\,\mathrm{m\,s^{-1}}. Find its greatest height above OO, using g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Plan. Take up positive and set v=0v=0 at the greatest height.

02=122+2(−9.8)suse the turning-point velocity19.6s=144rearranges=14419.6=7.3469… mdivide\begin{aligned} 0^2&=12^2+2(-9.8)s &&\text{use the turning-point velocity}\\ 19.6s&=144 &&\text{rearrange}\\ s&=\frac{144}{19.6}=7.3469\ldots\,\mathrm m &&\text{divide} \end{aligned}

Why use v=0v=0? The particle changes from rising to falling at that instant.

Answer. The greatest height above OO is 7.35 m7.35\,\mathrm m, to 33 significant figures.

Check. The time to the top is 12/9.812/9.8; substituting it into s=12t−4.9t2s=12t-4.9t^2 gives the same height.

Change only the launch speed to 16 m s−116\,\mathrm{m\,s^{-1}}. The method is unchanged, but the height does not grow in direct proportion to speed.

Worked example 6: a faster launch

Question. A particle is projected vertically upwards from OO at 16 m s−116\,\mathrm{m\,s^{-1}}. Find its greatest height above OO.

Plan. Use v=0v=0, upward positive and the same gravitational acceleration.

0=162−19.6sconstant-acceleration equations=25619.6=13.0612… msolve\begin{aligned} 0&=16^2-19.6s &&\text{constant-acceleration equation}\\ s&=\frac{256}{19.6}=13.0612\ldots\,\mathrm m &&\text{solve} \end{aligned}

Why square the speed? The equation relates a change of squared velocity to displacement.

Answer. 13.1 m13.1\,\mathrm m, to 33 significant figures.

Check. The ratio of heights is 256/144=16/9256/144=16/9, the square of the speed ratio 16/1216/12.

With downward positive, example 5 instead gives u=−12u=-12, a=9.8a=9.8 and s=−7.3469…s=-7.3469\ldots. The height above the launch point is still positive 7.35 m7.35\,\mathrm m.

Key idea

At the top, set vertical velocity to zero; keep gravitational acceleration. Use displacement from the stated initial point, with one consistent axis.

Find a new greatest height.

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