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M8.3

Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

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Weight and motion under gravity

Worked answers and methods for M8.3 on Edexcel A-level Maths 9MA0.

Explanation

  • Weight is the gravitational force W=mgW=mg, directed vertically downwards; mass is measured in kilograms and does not depend on location.
  • The value of gg is not universal and depends on location; use the stated value, or the Edexcel default 9.8 m s−29.8\,\text{m s}^{-2} when none is supplied.
  • For free motion near Earth's surface in the constant-gg model, acceleration is gg downwards, but a support or tension changes the resultant acceleration.
  • A common error is to call mass a force or to assume that the normal reaction always equals weight when the body has vertical acceleration.

Worked example

A particle is released from rest and falls freely through 19.6 m19.6\,\text{m}. Using g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the time taken and its speed after falling this distance.

  1. 1.Taking downwards as positive, s=ut+12gt2s=ut+\tfrac12gt^2 gives 19.6=0+4.9t219.6=0+4.9t^2, so t=2.0 st=2.0\,\text{s}.
  2. 2.Then v=u+gt=0+9.8(2)=19.6 m s−1v=u+gt=0+9.8(2)=19.6\,\text{m s}^{-1} downwards.

Answer: Time =2.0 s=2.0\,\text{s}; Speed =19.6 m s−1=19.6\,\text{m s}^{-1}

Common mistakes

  • Don't use g=9.8g=9.8 with distances in centimetres and times in seconds without converting units.
  • Don't use positive gravitational acceleration with an upward-positive displacement equation, producing inconsistent signs.

Exam tip

Declare the positive direction before substituting gravitational acceleration into a constant-acceleration equation.

Worked practice

Q1
Tier 1 · Easy

1.

Find the weight of a 2.5 kg2.5\,\text{kg} particle, using g=9.8 m s−2g=9.8\,\text{m s}^{-2}.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • 24.5 N24.5\,\text{N} vertically downwards
2
Notes
W=mg=2.5(9.8)=24.5 NW=mg=2.5(9.8)=24.5\,\text{N}, and weight acts vertically downwards.

(2 marks)

Q2
Tier 2 · Standard

2.

A particle is projected vertically upwards with speed 14 m s−114\,\text{m s}^{-1}. Using g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find its greatest height above the point of projection and the time taken to return to that point.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • Greatest height 10 m10\,\text{m}
  • Exact return time =207 s=\dfrac{20}{7}\,\text{s}
  • Return time to 33 significant figures =2.86 s=2.86\,\text{s}
4
Notes
At the greatest height, v=0v=0. Taking upwards as positive, v2=u2+2asv^2=u^2+2as gives 0=142−2(9.8)h0=14^2-2(9.8)h, so h=196/19.6=10 mh=196/19.6=10\,\text{m}. For the return to the starting level, 0=14t−4.9t20=14t-4.9t^2. The non-zero solution is t=14/4.9=20/7 st=14/4.9=20/7\,\text{s}.

(4 marks)

Q3
Tier 3 · Hard

3.

A person of mass 70 kg70\,\text{kg} stands on a scale in a lift. The scale exerts an upward force of 770 N770\,\text{N} on the person. Using g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude and direction of the lift's acceleration.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • 1.2 m s−21.2\,\text{m s}^{-2} upwards
4
Notes
The person's weight is 70(9.8)=686 N70(9.8)=686\,\text{N} downwards. Taking upwards as positive, Newton's second law gives 770−686=70a770-686=70a. Hence a=84/70=1.2 m s−2a=84/70=1.2\,\text{m s}^{-2} upwards.

(4 marks)

Q4
Tier 1 · Easy

4.

On a planet where g=3.7 m s−2g=3.7\,\text{m s}^{-2}, find the weight of a package of mass 6 kg6\,\text{kg}.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • 22.2 N22.2\,\text{N} vertically downwards
2
Notes
Weight is W=mg=6(3.7)=22.2 NW=mg=6(3.7)=22.2\,\text{N} and acts vertically downwards.

(2 marks)

Q5
Tier 2 · Standard

5.

A stone is projected vertically downwards at 3 m s−13\,\text{m s}^{-1} from a bridge. It falls 20 m20\,\text{m} before reaching the water. Using g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the time taken and the speed on reaching the water, giving both answers to 3 significant figures.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • Time =1.74 s=1.74\,\text{s}
  • Speed =20.0 m s−1=20.0\,\text{m s}^{-1}
4
Notes
Taking downwards as positive, 20=3t+4.9t220=3t+4.9t^2. The positive root is t=(−3+401)/9.8=1.74 st=(-3+\sqrt{401})/9.8=1.74\,\text{s} to 3 significant figures. Also v2=u2+2gs=32+2(9.8)(20)=401v^2=u^2+2gs=3^2+2(9.8)(20)=401, so v=401=20.0 m s−1v=\sqrt{401}=20.0\,\text{m s}^{-1} to 3 significant figures.

(4 marks)

Q6
Tier 3 · Hard

6.

On a moon, a probe is released from rest and falls 18 m18\,\text{m} in 2.4 s2.4\,\text{s}. Model the gravitational acceleration as constant. Find the value of gg on the moon and the weight there of equipment of mass 12 kg12\,\text{kg}. Given that g=9.8 m s−2g=9.8\,\text{m s}^{-2} on Earth, find the percentage by which the equipment's weight is lower on the moon. Give the percentage to 3 significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • Moon's g=6.25 m s−2g=6.25\,\text{m s}^{-2}
  • Moon weight =75 N=75\,\text{N}
  • Weight is 36.2%36.2\% lower
6
Notes
From s=12gt2s=\tfrac12gt^2, 18=12g(2.4)218=\tfrac12g(2.4)^2, so g=36/5.76=6.25 m s−2g=36/5.76=6.25\,\text{m s}^{-2}. The moon weight is 12(6.25)=75 N12(6.25)=75\,\text{N}; the Earth weight is 12(9.8)=117.6 N12(9.8)=117.6\,\text{N}. The percentage reduction is (117.6−75)/117.6×100=36.2%(117.6-75)/117.6\times100=36.2\% to 3 significant figures.

(6 marks)

Q7
Tier 2 · Standard

7.

A particle is released from rest and falls freely. Model it as a particle moving with constant acceleration and ignore air resistance. Take g=9.8 m s−2g=9.8\,\text{m s}^{-2}. Find the distance it falls during the third second of its motion and its speed after 3 s3\,\text{s}.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • Distance during the third second =24.5 m=24.5\,\text{m}
  • Speed after 3 s3\,\text{s} is 29.4 m s−129.4\,\text{m s}^{-1}
4
Notes
The distance fallen by time tt is s=12gt2=4.9t2s=\tfrac12gt^2=4.9t^2. The distance during the third second is s(3)−s(2)=4.9(9−4)=24.5 ms(3)-s(2)=4.9(9-4)=24.5\,\text{m}. The speed after 3 s3\,\text{s} is v=gt=9.8(3)=29.4 m s−1v=gt=9.8(3)=29.4\,\text{m s}^{-1}.

(4 marks)

Q8
Tier 3 · Hard

8.

On a planet, a particle is projected vertically upwards. Model the particle as moving under a constant gravitational acceleration of magnitude gg and ignore air resistance. The particle is at the same height after 1 s1\,\text{s} and after 3 s3\,\text{s}. At t=3 st=3\,\text{s} its velocity is 8 m s−18\,\text{m s}^{-1} vertically downwards. Find gg, the initial speed and the weight on this planet of an object of mass 4 kg4\,\text{kg}.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • g=8 m s−2g=8\,\text{m s}^{-2}
  • Initial speed =16 m s−1=16\,\text{m s}^{-1}
  • Weight =32 N=32\,\text{N}
6
Notes
Taking upwards as positive and writing the initial speed as uu, the height above the projection point is s=ut−12gt2s=ut-\tfrac12gt^2. Equality of the heights at t=1t=1 and t=3t=3 gives u−12g=3u−92gu-\tfrac12g=3u-\tfrac92g, so u=2gu=2g. Also v=u−gtv=u-gt, and at t=3t=3 the velocity is −8 m s−1-8\,\text{m s}^{-1}. Thus 2g−3g=−82g-3g=-8, giving g=8 m s−2g=8\,\text{m s}^{-2} and u=16 m s−1u=16\,\text{m s}^{-1}. The weight is mg=4(8)=32 Nmg=4(8)=32\,\text{N}.

(6 marks)

Q9
Tier 3 · Hard

9.

A package is released from a balloon that is moving vertically upwards at 14.7 m s−114.7\,\text{m s}^{-1}. At the instant of release the package is 19.6 m19.6\,\text{m} above horizontal ground. Model the package as a particle moving freely under gravity, ignore air resistance and take g=9.8 m s−2g=9.8\,\text{m s}^{-2}. Find the greatest height of the package above the ground, the time from release until it reaches the ground, and its velocity on impact. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • Greatest height =30.625 m=30.625\,\text{m}
  • Time to ground =4 s=4\,\text{s}
  • Impact velocity =24.5 m s−1=24.5\,\text{m s}^{-1} vertically downwards
6
Notes
Taking upwards as positive, at greatest height 0=14.72−2(9.8)h0=14.7^2-2(9.8)h, so the additional height is 11.025 m11.025\,\text{m} and the greatest height above ground is 19.6+11.025=30.625 m19.6+11.025=30.625\,\text{m}. Ground level satisfies 0=19.6+14.7t−4.9t20=19.6+14.7t-4.9t^2, or 0=4+3t−t20=4+3t-t^2, whose positive root is t=4t=4. The impact velocity is 14.7−9.8(4)=−24.5 m s−114.7-9.8(4)=-24.5\,\text{m s}^{-1}, namely 24.5 m s−124.5\,\text{m s}^{-1} downwards.

(6 marks)

Q10
Tier 3 · Hard

10.

Particle AA is released from rest at a point 63.7 m63.7\,\text{m} above horizontal ground. One second later, particle BB is projected vertically upwards from the ground at 19.6 m s−119.6\,\text{m s}^{-1}. Model both particles as moving freely under gravity, ignore air resistance and take g=9.8 m s−2g=9.8\,\text{m s}^{-2}. Find the time after AA is released when the particles meet, their height above the ground, and the velocity of each particle at that instant. Give all numerical answers exactly.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • Meeting time =3 s=3\,\text{s} after AA is released
  • Meeting height =19.6 m=19.6\,\text{m}
  • AA has velocity 29.4 m s−129.4\,\text{m s}^{-1} downwards
  • BB is instantaneously at rest
7
Notes
Let tt be the time after AA is released and take upwards as positive. Then yA=63.7−4.9t2y_A=63.7-4.9t^2. For t≥1t\geq1, yB=19.6(t−1)−4.9(t−1)2y_B=19.6(t-1)-4.9(t-1)^2. Equating and cancelling the common −4.9t2-4.9t^2 terms gives 63.7=29.4t−24.563.7=29.4t-24.5, so t=3t=3. The height is 63.7−4.9(9)=19.6 m63.7-4.9(9)=19.6\,\text{m}. The velocities are vA=−9.8(3)=−29.4 m s−1v_A=-9.8(3)=-29.4\,\text{m s}^{-1} and vB=19.6−9.8(3−1)=0v_B=19.6-9.8(3-1)=0.

(7 marks)

Verified exam appearances

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Other points in M8 Forces and Newton's laws

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