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A Level H2 Physics Mechanics Quiz
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A-Level Physics H2 Quiz – Mechanics: Answer Key
Total Marks: 80
Marking notes explain how marks are awarded. Partial credit may be given for correct method even if the final numerical answer is slightly off due to rounding. Always award method marks (M) and accuracy marks (A) as appropriate.
Section A (35 marks)
1. ( s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2.5 \times (8.0)^2 ) [M1]
( s = 0.5 \times 2.5 \times 64 = 80 ) m [A1]
Answer: 80 m
(3 marks: correct formula, substitution, answer with unit)
2.
(a) Use ( v^2 = u^2 - 2gs ) with ( v = 0 ) at max height: ( 0 = 15^2 - 2 \times 9.81 \times h ) [M1]
( h = \frac{225}{19.62} = 11.5 ) m (11.47 m) [A1]
2 marks
(b) Time to highest point ( t_{\text{up}} = \frac{u}{g} = \frac{15}{9.81} = 1.53 ) s [M1]
Total time ( = 2 \times 1.53 = 3.06 ) s [A1]
2 marks
(Total 4 marks)
3. Time of flight ( T = \frac{2u \sin\theta}{g} = \frac{2 \times 20 \times \sin 30^\circ}{9.81} = \frac{20}{9.81} = 2.039 ) s [M1]
Horizontal range ( R = u \cos\theta \times T = 20 \times \cos 30^\circ \times 2.039 ) [M1]
( R = 20 \times 0.8660 \times 2.039 = 35.3 ) m [A1]
3 marks: time of flight, range expression, final value with unit
4.
(a) Normal reaction ( N = mg = 5.0 \times 9.81 = 49.05 ) N [M1]
Maximum static friction ( f_{\text{s,max}} = \mu_s N = 0.40 \times 49.05 = 19.62 ) N.
Since applied force 25 N > 19.62 N, the block moves. [A1]
2 marks
(b) Kinetic friction ( f_k = \mu_k N = 0.30 \times 49.05 = 14.7 ) N [M1]
Net force ( F_{\text{net}} = 25 - 14.7 = 10.3 ) N [M1]
Acceleration ( a = \frac{F_{\text{net}}}{m} = \frac{10.3}{5.0} = 2.06 ) m s⁻² [A1]
(3 marks: kinetic friction, net force, acceleration)
Answers: (a) moves; (b) acceleration = 2.06 m s⁻², friction = 14.7 N
5.
(a) Vertical equilibrium: ( T \cos 30^\circ = mg ) [M1]
( T = \frac{2.0 \times 9.81}{\cos 30^\circ} = \frac{19.62}{0.8660} = 22.6 ) N [A1]
1.5 marks (method, answer)
(b) Horizontal component provides centripetal force: ( T \sin 30^\circ = \frac{mv^2}{r} ) where ( r = L \sin 30^\circ = 0.40 ) m [M1]
( v = \sqrt{\frac{T \sin 30^\circ \times r}{m}} = \sqrt{\frac{22.6 \times 0.5 \times 0.40}{2.0}} = \sqrt{2.26} = 1.50 ) m s⁻¹ [A1]
1.5 marks
(Total 3 marks)
6. Orbital radius ( r = 6.37 \times 10^6 + 0.30 \times 10^6 = 6.67 \times 10^6 ) m
(a) ( v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67\times10^{-11} \times 5.97\times10^{24}}{6.67\times10^6}} ) [M1]
( = \sqrt{5.97 \times 10^7} = 7.73 \times 10^3 ) m s⁻¹ [A1]
2 marks
(b) Period ( T = \frac{2\pi r}{v} = \frac{2\pi \times 6.67\times10^6}{7.73\times10^3} ) [M1]
( = 5.42 \times 10^3 ) s = ( 90.4 ) min (or 90 min) [A1]
2 marks
(Total 4 marks)
7. For elastic collision with stationary target:
( v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 = \frac{0.50 - 1.0}{1.5}\times 4.0 = -\frac{0.5}{1.5}\times 4.0 = -1.33 ) m s⁻¹ [M1]
( v_2 = \frac{2m_1}{m_1 + m_2}u_1 = \frac{2\times0.50}{1.5}\times 4.0 = \frac{1.0}{1.5}\times 4.0 = 2.67 ) m s⁻¹ [A1]
3 marks for (a): correct formula application, both velocities
(b) Impulse on 0.50 kg cart = change in momentum = ( m(v - u) = 0.50(-1.33 - 4.0) = 0.50 \times (-5.33) = -2.67 ) N s [M1]
Magnitude = 2.67 N s, direction opposite to initial motion. [A1]
(Total 5 marks)
8. For no friction, banking angle ( \theta ) satisfies ( \tan \theta = \frac{v^2}{rg} ) [M1]
( \tan \theta = \frac{20^2}{50 \times 9.81} = \frac{400}{490.5} = 0.8155 ) [M1]
( \theta = \arctan(0.8155) = 39.2^\circ ) (accept ( 39^\circ)) [A1]
3 marks: formula, substitution, answer; total 3 marks as indicated (question states 4 marks? Actually the text said 4 marks originally but I'll adjust to 4 marks by adding an additional step: explain condition. Let’s give 1 mark for stating condition and 3 for calculation → 4 marks.)
Award: stating correct condition (tanθ = v²/rg) [1]; substitution correct [1]; answer with unit [1]; correct justification that this gives zero friction demand [1]. So:
Banking angle = 39.2°, marks as above.
Total 4 marks
9. The principle of conservation of linear momentum states that in a closed (or isolated) system (where no external forces act), the total momentum before an event is equal to the total momentum after the event. [M1, A1]
2 marks: mention of closed/isolated system, equality of before/after total momentum
10.
(a) Speed before impact: ( v_1 = \sqrt{2gh_1} = \sqrt{2\times9.81\times1.8} = 5.94 ) m s⁻¹ (downward) [A1]
(b) Speed just after impact: ( v_2 = \sqrt{2gh_2} = \sqrt{2\times9.81\times1.2} = 4.85 ) m s⁻¹ (upward) [A1]
(c) Impulse = change in momentum ( \Delta p = m(v_f - v_i) ) taking upward as positive: ( \Delta p = 0.20(4.85 - (-5.94)) = 0.20 \times 10.79 = 2.16 ) N s upward. [A1]
Total 3 marks
Section B (45 marks)
11.
(a) ( a = \frac{\Delta v}{t} = \frac{25 - 0}{10} = 2.5 ) m s⁻² [A1]
(b) ( s = \frac{1}{2}at^2 = \frac{1}{2} \times 2.5 \times 100 = 125 ) m [A1]
(c) Method 1: ( \text{KE gained} = \frac{1}{2}mv^2 = \frac{1}{2}\times1500\times25^2 = 468750 ) J; average power ( = \frac{\text{KE}}{t} = 46875 ) W [M1, A1]
or Method 2: force ( F = ma = 3750 ) N, work done ( = Fs = 468750 ) J, power = work/time = 46875 W.
2 marks for (c): any valid method.
(Total 4 marks)
12.
(a) Net accelerating force = weight of ( m_2 = m_2 g = 2.0 \times 9.81 = 19.62 ) N [M1]
Total mass accelerated = ( 5.0 ) kg. ( a = \frac{19.62}{5.0} = 3.92 ) m s⁻² [A1]
2 marks
(b) Consider ( m_1 ): ( T = m_1 a = 3.0 \times 3.92 = 11.8 ) N [M1, A1]
or ( m_2 ): ( m_2 g - T = m_2 a \Rightarrow T = 19.62 - 2.0 \times 3.92 = 11.8 ) N.
2 marks
(c) Use ( v^2 = u^2 + 2as ), ( u=0 ): ( v = \sqrt{2 \times 3.92 \times 1.5} = \sqrt{11.76} = 3.43 ) m s⁻¹ [A1]
1 mark
(Total 5 marks)
13.
( v = 10t - 2t^2 )
(a) ( a = \frac{dv}{dt} = 10 - 4t ). At ( t=0 ), ( a = 10 ) m s⁻². [A1]
(b) Maximum velocity occurs when ( a = 0 ): ( 10 - 4t = 0 \Rightarrow t = 2.5 ) s. [A1]
(c) Displacement ( s = \int_0^5 (10t - 2t^2) dt = \left[5t^2 - \frac{2}{3}t^3\right]_0^5 = 5(25) - \frac{2}{3}(125) = 125 - 83.3 = 41.7 ) m. [A1]
(Total 3 marks)
14.
At top: ( T_{\text{top}} + mg = \frac{mv^2}{r} )
( T_{\text{top}} = \frac{0.05 \times 16}{0.80} - 0.05 \times 9.81 = 1.00 - 0.4905 = 0.510 ) N [M1, A1]
At bottom: ( T_{\text{bottom}} - mg = \frac{mv^2}{r} )
( T_{\text{bottom}} = 1.00 + 0.4905 = 1.49 ) N [M1, A1]
(Total 4 marks)
15.
(a) Conservation of momentum: ( 1000 \times 15 = (1000 + 2000) v )
( v = \frac{15000}{3000} = 5.0 ) m s⁻¹ [M1, A1]
2 marks
(b) Initial KE = ( \frac{1}{2} \times 1000 \times 15^2 = 112500 ) J
Final KE = ( \frac{1}{2} \times 3000 \times 5^2 = 37500 ) J
Loss = ( 112500 - 37500 = 75000 ) J [M1, A1]
2 marks
(c) The lost kinetic energy is converted into heat, sound, and deformation of the vehicles during the collision. [A1]
(Total 5 marks)
16. Gravitational field strength ( g = \frac{GM}{R^2} )
For planet: ( g' = \frac{G(2M)}{(1.5R)^2} = \frac{2GM}{2.25R^2} = \frac{2}{2.25} \frac{GM}{R^2} = 0.889,g ) [M2, A1]
3 marks: correct ratio, simplification, final expression
17.
(a) Component down slope = ( mg \sin 20^\circ = 70 \times 9.81 \times 0.342 = 235 ) N [M1, A1]
2 marks
(b) Using energy: loss in PE = gain in KE: ( mgL\sin 20^\circ = \frac{1}{2}mv^2 )
( v = \sqrt{2gL\sin 20^\circ} = \sqrt{2 \times 9.81 \times 50 \times 0.342} = \sqrt{335.5} = 18.3 ) m s⁻¹ [M1, A1]
2 marks
(Total 4 marks)
18. Change in momentum: ( \Delta p = m(v_f - v_i) = 0.15(50 - (-40)) = 0.15 \times 90 = 13.5 ) kg m s⁻¹ [M1]
Average force: ( F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{13.5}{2.0 \times 10^{-3}} = 6750 ) N [M1, A1]
3 marks
19. Mass ( m = 0.500 ) kg, ( k = 20 ) N m⁻¹, amplitude ( A = 0.10 ) m.
(a) Maximum speed occurs at equilibrium: ( \frac{1}{2}kA^2 = \frac{1}{2}mv_{\max}^2 \Rightarrow v_{\max} = A\sqrt{\frac{k}{m}} = 0.10 \sqrt{\frac{20}{0.50}} = 0.632 ) m s⁻¹ [M1, A1]
2 marks
(b) At ( x = 0.05 ) m: total energy ( E = \frac{1}{2}kA^2 = 0.10 ) J, PE ( = \frac{1}{2}kx^2 = \frac{1}{2}\times20\times(0.05)^2 = 0.025 ) J.
KE = 0.10 – 0.025 = 0.075 J; ( v = \sqrt{\frac{2\times0.075}{0.50}} = 0.548 ) m s⁻¹ [M1, A1]
2 marks
(Total 4 marks)
20.
(a) For equilibrium: ( m_A g d_A = m_B g d_B )
( 30 \times 9.81 \times 2.0 = 20 \times 9.81 \times d_B )
( d_B = \frac{60}{20} = 3.0 ) m (cancelling ( g )) [M1, A1]
2 marks
(b) New torque from Child B: ( 20 \times 9.81 \times (3.0 + 0.5) = 20 \times 9.81 \times 3.5 = 686.7 ) N m [M1]
Torque from Child A: ( 30 \times 9.81 \times 2.0 = 588.6 ) N m
Extra torque needed on left side to balance: ( 686.7 - 588.6 = 98.1 ) N m [M1, A1]
(Accept 98.0 N m)
3 marks
(Total 5 marks)
End of Answer Key