Hard A-Level Forces and Newton’s laws Questions

Challenging, exam-style A-Level Forces and Newton’s laws questions with worked solutions. Stretch yourself on the hardest resultant, magnitude, direction, newtons-second-law problems.

resultantmagnitudedirectionnewtons-second-lawvectorsequilibrium
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A particle is in equilibrium under three forces. Two are (7i−3j) N(7\mathbf{i}-3\mathbf{j})\,\text{N} and (−2i+5j) N(-2\mathbf{i}+5\mathbf{j})\,\text{N}. Which is the magnitude of the third force?
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Worked solution

  1. State the equilibrium condition

    F1+F2+F3=0\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}

    The forces sum to zero.

  2. Add the i-components of the known forces

    Σx=7+(−2)=5\Sigma_x=7+(-2)=5

    Sum the horizontal components.

  3. Add the j-components of the known forces

    Σy=−3+5=2\Sigma_y=-3+5=2

    Sum the vertical components.

  4. Write the resultant of the known forces

    F1+F2=(5i+2j) N\mathbf{F}_1+\mathbf{F}_2=(5\mathbf{i}+2\mathbf{j})\,\text{N}

    Combine the two known forces.

  5. The third force cancels this

    F3=−(5i+2j)\mathbf{F}_3=-(5\mathbf{i}+2\mathbf{j})

    Equal and opposite to the resultant.

  6. Write its components

    F3=(−5i−2j) N\mathbf{F}_3=(-5\mathbf{i}-2\mathbf{j})\,\text{N}

    Negate each component.

  7. Write the magnitude formula

    ∣F3∣=(−5)2+(−2)2|\mathbf{F}_3|=\sqrt{(-5)^{2}+(-2)^{2}}

    Use Pythagoras.

  8. Simplify under the root

    ∣F3∣=29|\mathbf{F}_3|=\sqrt{29}

    Add the squares.

  9. Evaluate the magnitude

    ∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}

    The size of the third force.

  10. Verify the sum is zero

    (5−5)i+(2−2)j=0(5-5)\mathbf{i}+(2-2)\mathbf{j}=\mathbf{0}

    The three forces balance.

  11. Note the magnitude ignores sign

    ∣F3∣=∣−(5i+2j)∣|\mathbf{F}_3|=|{-}(5\mathbf{i}+2\mathbf{j})|

    Magnitude of a vector and its negative are equal.

  12. Eliminate the sum-of-magnitudes value

    ∣F3∣≠7.62|\mathbf{F}_3|\neq 7.62

    Do not add the magnitudes of the two forces.

  13. Eliminate the component values

    ∣F3∣≠5|\mathbf{F}_3|\neq 5

    5 is only the i-component.

  14. Check the arithmetic

    29=5.39\sqrt{29}=5.39

    Confirms the magnitude.

  15. Select the correct value

    ∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}

    This is the required magnitude to 3 s.f.

Answer
∣F3∣=5.39 N|\mathbf{F}_3|=5.39\,\text{N}
Question 2
8 markschallenging
A person of mass 70 kg70\,\text{kg} stands in a lift that accelerates upward at 2 m s−22\,\text{m s}^{-2}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which is the force exerted by the floor on the person?
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Worked solution

  1. Model the forces

    R up, W=mg downR\ \text{up},\ W=mg\ \text{down}

    Floor reaction up, weight down.

  2. Write the weight

    W=70×9.8W=70\times 9.8

    Weight is mass times g.

  3. Evaluate the weight

    W=686 NW=686\,\text{N}

    The person's true weight.

  4. State the acceleration

    a=2 m s−2 upa=2\,\text{m s}^{-2}\ \text{up}

    The lift accelerates upward.

  5. Apply Newton's second law upward

    R−mg=maR-mg=ma

    Take upward as positive.

  6. Rearrange for the reaction

    R=m(g+a)R=m(g+a)

    Make R the subject.

  7. Substitute the values

    R=70(9.8+2)R=70(9.8+2)

    Insert m, g and a.

  8. Simplify inside the bracket

    R=70×11.8R=70\times 11.8

    Add g and a.

  9. Evaluate the reaction

    R=826 NR=826\,\text{N}

    The floor pushes with this force.

  10. Compare with the weight

    R>WR>W

    The reaction exceeds the weight.

  11. Interpret physically

    feels heavier\text{feels heavier}

    The passenger feels heavier.

  12. Eliminate the weight value

    R≠686R\neq 686

    That would need constant velocity.

  13. Eliminate the decelerating value

    R≠546R\neq 546

    That corresponds to slowing down.

  14. Check the arithmetic

    70×11.8=82670\times 11.8=826

    Confirms the reaction.

  15. Select the correct value

    R=826 NR=826\,\text{N}

    This is the floor reaction.

Answer
R=826 NR=826\,\text{N}
Question 3
8 markschallenging
Two forces (5i+2j) N(5\mathbf{i}+2\mathbf{j})\,\text{N} and (−i+4j) N(-\mathbf{i}+4\mathbf{j})\,\text{N} act on a particle of mass 2 kg2\,\text{kg}. Which statement about the resulting acceleration is correct?
Show worked solution

Worked solution

  1. State Newton's second law

    ∑F=ma\sum\mathbf{F}=m\mathbf{a}

    Resultant force equals mass times acceleration.

  2. Add the i-components

    Rx=5+(−1)=4R_x=5+(-1)=4

    Sum the horizontal components.

  3. Add the j-components

    Ry=2+4=6R_y=2+4=6

    Sum the vertical components.

  4. Write the resultant force

    R=(4i+6j) N\mathbf{R}=(4\mathbf{i}+6\mathbf{j})\,\text{N}

    The total force on the particle.

  5. Write the magnitude formula

    ∣R∣=42+62|\mathbf{R}|=\sqrt{4^{2}+6^{2}}

    Use Pythagoras.

  6. Simplify under the root

    ∣R∣=52|\mathbf{R}|=\sqrt{52}

    Add the squares.

  7. Evaluate the force magnitude

    ∣R∣=7.21 N|\mathbf{R}|=7.21\,\text{N}

    The size of the resultant.

  8. Find the acceleration vector

    a=12(4i+6j)\mathbf{a}=\tfrac12(4\mathbf{i}+6\mathbf{j})

    Divide by the mass.

  9. Evaluate the acceleration components

    a=(2i+3j) m s−2\mathbf{a}=(2\mathbf{i}+3\mathbf{j})\,\text{m s}^{-2}

    Halve each component.

  10. Write the acceleration magnitude

    ∣a∣=22+32|\mathbf{a}|=\sqrt{2^{2}+3^{2}}

    Use Pythagoras on the acceleration.

  11. Simplify

    ∣a∣=13|\mathbf{a}|=\sqrt{13}

    Add the squares.

  12. Evaluate the magnitude

    ∣a∣=3.61 m s−2|\mathbf{a}|=3.61\,\text{m s}^{-2}

    The size of the acceleration.

  13. Check against the force magnitude

    ∣a∣=∣R∣2|\mathbf{a}|=\frac{|\mathbf{R}|}{2}

    Consistent with dividing by the mass.

  14. Eliminate the incorrect values

    ∣a∣≠7.21|\mathbf{a}|\neq 7.21

    That is the force magnitude, not the acceleration.

  15. Select the correct statement

    ∣a∣=3.61 m s−2|\mathbf{a}|=3.61\,\text{m s}^{-2}

    This is the acceleration magnitude to 3 s.f.

Answer
∣a∣=3.61 m s−2|\mathbf{a}|=3.61\,\text{m s}^{-2}
Question 4
8 markschallenging
A particle is held in equilibrium by three forces. Two of them are F1=(9i+2j) N\mathbf{F}_1=(9\mathbf{i}+2\mathbf{j})\,\text{N} and F2=(−3i−8j) N\mathbf{F}_2=(-3\mathbf{i}-8\mathbf{j})\,\text{N}. Find the magnitude of the third force and the angle it makes with the i\mathbf{i}-direction, to 3 significant figures.
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Worked solution

  1. State the condition for equilibrium

    F1+F2+F3=0\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}

    In equilibrium the vector sum of the forces is zero.

  2. List the known forces

    F1=(9i+2j), F2=(−3i−8j)\mathbf{F}_1=(9\mathbf{i}+2\mathbf{j}),\ \mathbf{F}_2=(-3\mathbf{i}-8\mathbf{j})

    Two of the three forces are given.

  3. Add the i-components of the known forces

    Σx=9+(−3)=6\Sigma_x=9+(-3)=6

    Sum the horizontal components.

  4. Add the j-components of the known forces

    Σy=2+(−8)=−6\Sigma_y=2+(-8)=-6

    Sum the vertical components.

  5. Write the resultant of the known forces

    F1+F2=(6i−6j) N\mathbf{F}_1+\mathbf{F}_2=(6\mathbf{i}-6\mathbf{j})\,\text{N}

    Combine the two known forces.

  6. The third force cancels this

    F3=−(F1+F2)\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)

    F_3 is equal and opposite to their resultant.

  7. Write the components of the third force

    F3=(−6i+6j) N\mathbf{F}_3=(-6\mathbf{i}+6\mathbf{j})\,\text{N}

    Negate each component.

  8. Write the magnitude formula

    ∣F3∣=Fx2+Fy2|\mathbf{F}_3|=\sqrt{F_x^{2}+F_y^{2}}

    Use Pythagoras on its components.

  9. Substitute the components

    ∣F3∣=−62+62|\mathbf{F}_3|=\sqrt{-6^{2}+6^{2}}

    Insert the two components.

  10. Evaluate the magnitude

    ∣F3∣=8.49 N|\mathbf{F}_3|=8.49\,\text{N}

    The size of the balancing force.

  11. Set up the direction

    tan⁡θ=6−6\tan\theta=\frac{6}{-6}

    The angle of the third force.

  12. Evaluate the direction

    θ=135∘\theta=135^\circ

    Measured from the i-direction (taking account of the quadrant).

  13. Verify the forces sum to zero

    F1+F2+F3=0\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}

    Adding all three returns the zero vector.

  14. Confirm the units

    ∣F3∣ is in newtons|\mathbf{F}_3|\ \text{is in newtons}

    The third force is a vector force.

  15. State the magnitude and direction

    ∣F3∣=8.49 N, θ=135∘|\mathbf{F}_3|=8.49\,\text{N},\ \theta=135^\circ

    This is the required force to 3 s.f.

Answer
∣F3∣=8.49 N, θ=135∘|\mathbf{F}_3|=8.49\,\text{N},\ \theta=135^\circ
Question 5
8 markschallenging
A particle of mass 4 kg4\,\text{kg} is initially at rest. Two constant forces F1=(6i+8j) N\mathbf{F}_1=(6\mathbf{i}+8\mathbf{j})\,\text{N} and F2=(4i−2j) N\mathbf{F}_2=(4\mathbf{i}-2\mathbf{j})\,\text{N} act on it. Find its speed and the distance it travels after 8 s8\,\text{s}. Give each answer to 3 significant figures.
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Worked solution

  1. State Newton's second law

    ∑F=ma\sum\mathbf{F}=m\mathbf{a}

    The resultant force gives the acceleration.

  2. List the two forces

    F1=(6i+8j), F2=(4i−2j)\mathbf{F}_1=(6\mathbf{i}+8\mathbf{j}),\ \mathbf{F}_2=(4\mathbf{i}-2\mathbf{j})

    Both forces are given in components.

  3. Add the i-components

    Rx=6+(4)=10R_x=6+(4)=10

    Sum the horizontal components.

  4. Add the j-components

    Ry=8+(−2)=6R_y=8+(-2)=6

    Sum the vertical components.

  5. Write the resultant force

    R=(10i+6j) N\mathbf{R}=(10\mathbf{i}+6\mathbf{j})\,\text{N}

    The total force on the particle.

  6. Write the magnitude formula

    ∣R∣=102+62|\mathbf{R}|=\sqrt{10^{2}+6^{2}}

    Use Pythagoras on the components.

  7. Evaluate the magnitude of the force

    ∣R∣=11.7 N|\mathbf{R}|=11.7\,\text{N}

    The size of the resultant force.

  8. Find the magnitude of the acceleration

    ∣a∣=∣R∣m=11.74|\mathbf{a}|=\frac{|\mathbf{R}|}{m}=\frac{11.7}{4}

    Divide the force magnitude by the mass.

  9. Evaluate the acceleration

    ∣a∣=2.92 m s−2|\mathbf{a}|=2.92\,\text{m s}^{-2}

    Carry out the division.

  10. Find the direction of motion

    θ=31∘\theta=31^\circ

    Motion is along the resultant force.

  11. The particle starts from rest

    u=0 m s−1u=0\,\text{m s}^{-1}

    It begins at rest, so u=0.

  12. Find the speed with v=u+at

    v=∣a∣t=2.92×8v=|\mathbf{a}|t=2.92\times 8

    Speed after the given time.

  13. Evaluate the speed

    v=23.3 m s−1v=23.3\,\text{m s}^{-1}

    Compute the product.

  14. Find the distance with s=\tfrac12 at^{2}

    s=12×2.92×82=93.3 ms=\tfrac12\times 2.92\times 8^{2}=93.3\,\text{m}

    Displacement in the given time.

  15. State the results

    v=23.3 m s−1, s=93.3 mv=23.3\,\text{m s}^{-1},\ s=93.3\,\text{m}

    These are the required speed and distance.

Answer
v=23.3 m s−1, s=93.3 mv=23.3\,\text{m s}^{-1},\ s=93.3\,\text{m}

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