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A Level H1 Physics Practice Paper 3

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TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

Answer Key and Marking Scheme
Paper: Practice Paper (Version 3 of 5)
Topic: Mechanics

Section A: Structured Questions

1.
(a) Linear momentum is the product of mass and velocity. [B1]
(b) p=mv=0.85×1.2p = mv = 0.85 \times 1.2 [M1]
p=1.02 kg m s−1p = 1.02 \text{ kg m s}^{-1} [A1]
(c) Newton’s First Law states that an object remains at rest or in uniform motion unless acted upon by a net external force. [B1]
Since the velocity is constant (zero acceleration), the net force must be zero. [B1]

2.
(a) Forces:

  1. Weight of plank (120 N120 \text{ N}) acting downwards at the center (2.0 m2.0 \text{ m} from A). [B1]
  2. Weight of student (500 N500 \text{ N}) acting downwards at distance xx. [B1]
  3. Reaction force at X (RXR_X) acting upwards at A. [B1]
  4. Reaction force at Y (RYR_Y) acting upwards at 3.0 m3.0 \text{ m} from A. [B1]
    (Note: Accept arrows labeled clearly. Max 3 marks if only 3 forces shown but correct.)
    (b) At the point of tipping, the plank loses contact with support X, so RX=0R_X = 0. [B1]
    Take moments about support Y (position 3.0 m3.0 \text{ m} from A).
    Clockwise moment = Anti-clockwise moment.
    Weight of plank acts at 2.0 m2.0 \text{ m} from A. Distance from Y = 3.0−2.0=1.0 m3.0 - 2.0 = 1.0 \text{ m}.
    Moment of plank = 120×1.0=120 N m120 \times 1.0 = 120 \text{ N m}. [M1]
    Student is at distance xx from A. Distance from Y = 3.0−x3.0 - x.
    Moment of student = 500×(3.0−x)500 \times (3.0 - x). [M1]
    Equilibrium: 500(3.0−x)=120500(3.0 - x) = 120
    1500−500x=1201500 - 500x = 120
    500x=1380500x = 1380
    x=2.76 mx = 2.76 \text{ m} [A1]

3.
(a) In a closed/isolated system (no external forces), the total linear momentum before interaction equals the total linear momentum after interaction. [B1] [B1]
(b) Total initial momentum = 00.
Let right be positive.
pinitial=pfinalp_{\text{initial}} = p_{\text{final}}
0=mPvP+mQvQ0 = m_P v_P + m_Q v_Q
0=60(−2.5)+80(vQ)0 = 60(-2.5) + 80(v_Q) [M1]
150=80vQ150 = 80 v_Q
vQ=1.875 m s−1v_Q = 1.875 \text{ m s}^{-1} [A1]
Direction: To the right. [B1]
(c) KEP=12(60)(2.5)2=187.5 JKE_P = \frac{1}{2}(60)(2.5)^2 = 187.5 \text{ J} [M1]
KEQ=12(80)(1.875)2=140.625 JKE_Q = \frac{1}{2}(80)(1.875)^2 = 140.625 \text{ J} [M1]
Total KE=187.5+140.625=328.125 JKE = 187.5 + 140.625 = 328.125 \text{ J} (approx 328 J328 \text{ J}) [A1]

4.
(a) v2=u2+2asv^2 = u^2 + 2as
At max height, v=0v = 0. u=15u = 15, a=−9.81a = -9.81.
0=152+2(−9.81)s0 = 15^2 + 2(-9.81)s [M1]
19.62s=22519.62 s = 225
s=11.47 ms = 11.47 \text{ m} (approx 11.5 m11.5 \text{ m}) [A1]
(b) Graph: Straight line with negative gradient. [B1]
Starts at positive vv (+15+15), crosses t-axis (v=0), ends at negative vv (−15-15). [B1]
(c) Air resistance opposes motion.
On the way up, air resistance acts downwards (with gravity), so deceleration >g> g. Time to top is shorter. [B1]
On the way down, air resistance acts upwards (against gravity), so acceleration <g< g. Time to fall is longer. [B1]

5.
(a) Component of weight down slope = mgsin⁡θmg \sin \theta
=1200×9.81×sin⁡(5.0∘)= 1200 \times 9.81 \times \sin(5.0^\circ) [M1]
=1025 N= 1025 \text{ N} (approx 1030 N1030 \text{ N}) [A1]
(b) Since speed is constant, driving force FDF_D balances resistive forces.
FD=Weight component+Resistive forceF_D = \text{Weight component} + \text{Resistive force}
FD=1025+400=1425 NF_D = 1025 + 400 = 1425 \text{ N} [M1]
Power P=FDvP = F_D v
P=1425×20P = 1425 \times 20 [M1]
P=28,500 WP = 28,500 \text{ W} (or 28.5 kW28.5 \text{ kW}) [A1]

Section B: Data and Context Questions

6.
(a) h=12gt2h = \frac{1}{2}gt^2 is in the form y=mxy = mx where y=hy=h, x=t2x=t^2, and m=12gm = \frac{1}{2}g. [B1]
Since there is no constant term (c=0c=0), the line passes through the origin. [B1]
(b) Plot points correctly. [B1]
Line of best fit is a straight line through the origin. [B1]
Axes labeled with units. [B1]
Scale appropriate. [B1]
(c) Gradient = ΔhΔt2\frac{\Delta h}{\Delta t^2}. Using points from line (e.g., (0.5,2.5)(0.5, 2.5)):
Gradient ≈2.50.5=5.0 m s−2\approx \frac{2.5}{0.5} = 5.0 \text{ m s}^{-2} (Accept range 4.8−5.24.8 - 5.2). [M1] [A1]
(d) Gradient =12g⇒g=2×Gradient= \frac{1}{2}g \Rightarrow g = 2 \times \text{Gradient}.
g=2×5.0=10.0 m s−2g = 2 \times 5.0 = 10.0 \text{ m s}^{-2} (Accept range 9.6−10.49.6 - 10.4). [M1] [A1]

7.
(a) Loss in GPE = Gain in KE
mgh=12mv2mgh = \frac{1}{2}mv^2
v=2gh=2×9.81×0.80v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.80} [M1]
v=15.696=3.96 m s−1v = \sqrt{15.696} = 3.96 \text{ m s}^{-1} [A1]
(b) Vertical motion: s=ut+12at2s = ut + \frac{1}{2}at^2. uy=0u_y = 0, s=0.45s = 0.45, a=9.81a = 9.81.
0.45=0+12(9.81)t20.45 = 0 + \frac{1}{2}(9.81)t^2
t2=0.909.81⇒t=0.303 st^2 = \frac{0.90}{9.81} \Rightarrow t = 0.303 \text{ s} [M1]
Horizontal motion: sx=vxts_x = v_x t.
1.2=vx(0.303)1.2 = v_x (0.303)
vx=1.20.303=3.96 m s−1v_x = \frac{1.2}{0.303} = 3.96 \text{ m s}^{-1} [M1] [A1]
(Note: If student uses g=10g=10, answers will vary slightly but method marks apply.)
(c) Energy lost due to friction/resistance on the track. [B1]

8.
(a) k=Fx=200.10=200 N m−1k = \frac{F}{x} = \frac{20}{0.10} = 200 \text{ N m}^{-1}. [M1] [A1]
(b) EPE=12kx2=12(200)(0.10)2EPE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2 [M1]
EPE=1.0 JEPE = 1.0 \text{ J}. [A1]
(c) EPE=GPEEPE = GPE
1.0=mgh1.0 = mgh
1.0=0.05×9.81×h1.0 = 0.05 \times 9.81 \times h [M1]
h=1.00.4905=2.04 mh = \frac{1.0}{0.4905} = 2.04 \text{ m}. [A1]

9.
(a) Conservation of momentum:
mAuA+mBuB=(mA+mB)vm_A u_A + m_B u_B = (m_A + m_B)v
2.0(3.0)+1.0(0)=(2.0+1.0)v2.0(3.0) + 1.0(0) = (2.0 + 1.0)v [M1]
6.0=3.0v6.0 = 3.0v
v=2.0 m s−1v = 2.0 \text{ m s}^{-1}. [A1] Direction: Right. [B1]
(b) KEinitial=12(2.0)(3.0)2=9.0 JKE_{\text{initial}} = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}. [M1]
KEfinal=12(3.0)(2.0)2=6.0 JKE_{\text{final}} = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}. [M1]
KEinitial≠KEfinalKE_{\text{initial}} \neq KE_{\text{final}} (9.0≠6.09.0 \neq 6.0), so kinetic energy is not conserved. [B1]
Therefore, the collision is inelastic. [B1]

10.
(a) Fnet=ma=5.0×2.0=10 NF_{\text{net}} = ma = 5.0 \times 2.0 = 10 \text{ N}. [M1] [A1]
(b) Fnet=Fpull−FfrictionF_{\text{net}} = F_{\text{pull}} - F_{\text{friction}}
10=30−Ffriction10 = 30 - F_{\text{friction}} [M1]
Ffriction=20 NF_{\text{friction}} = 20 \text{ N}. [A1]
(c) The block will decelerate (slow down) due to the frictional force acting opposite to motion. [B1]
It will eventually come to rest. [B1]

Section C: Extended Response

11.
(a) Initially, air resistance is zero, so net force is weight (WW). Acceleration is gg. [B1]
As speed increases, air resistance (RR) increases. Net force =W−R= W - R. [B1]
Since net force decreases, acceleration decreases. [B1]
When R=WR = W, net force is zero, acceleration is zero, and terminal velocity is reached. [B1]
(b) Opening parachute greatly increases surface area, causing a large increase in air resistance. [B1]
Air resistance becomes much greater than weight (R>WR > W). [B1]
This creates a large net upward force, causing rapid deceleration (upward acceleration). [B1]
(c) Air resistance equals weight (R=WR = W). [B1]

12.
(a) Forces: Tension TT (up), Weight mgmg (down).
Fnet=T−mg=maF_{\text{net}} = T - mg = ma [M1]
T=m(g+a)T = m(g + a)
T=500(9.81+0.50)T = 500(9.81 + 0.50) [M1]
T=500(10.31)=5155 NT = 500(10.31) = 5155 \text{ N}. [A1] (Accept 51505150 or 51605160)
(b) Distance moved s=ut+12at2=0+12(0.50)(4.0)2=4.0 ms = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ m}. [M1]
Work Done =T×s= T \times s [M1]
W=5155×4.0=20,620 JW = 5155 \times 4.0 = 20,620 \text{ J}. [A1] (Accept 20,60020,600)
(c) Average Power =WorkTime= \frac{\text{Work}}{\text{Time}} [M1]
P=206204.0=5155 WP = \frac{20620}{4.0} = 5155 \text{ W}. [A1]

13.
(a) Take direction towards wall as positive.
u=+12 m s−1u = +12 \text{ m s}^{-1}, v=−10 m s−1v = -10 \text{ m s}^{-1}.
Δp=m(v−u)=0.15(−10−12)\Delta p = m(v - u) = 0.15(-10 - 12) [M1]
Δp=0.15(−22)=−3.3 kg m s−1\Delta p = 0.15(-22) = -3.3 \text{ kg m s}^{-1}. [A1]
Magnitude: 3.3 kg m s−13.3 \text{ kg m s}^{-1}. Direction: Away from wall. [B1]
(b) Favg=ΔpΔtF_{\text{avg}} = \frac{\Delta p}{\Delta t} [M1]
F=3.30.02=165 NF = \frac{3.3}{0.02} = 165 \text{ N}. [A1]
(c) KEinitial=12(0.15)(12)2=10.8 JKE_{\text{initial}} = \frac{1}{2}(0.15)(12)^2 = 10.8 \text{ J}.
KEfinal=12(0.15)(10)2=7.5 JKE_{\text{final}} = \frac{1}{2}(0.15)(10)^2 = 7.5 \text{ J}.
KE is lost to sound, heat, and deformation of the ball/wall. [B1] [B1]

14.
(a) The wall is smooth, so there is no friction. The reaction force must be perpendicular (normal) to the surface. [B1]
(b) Diagram:

  1. Weight WW down from center. [B1]
  2. Normal reaction from ground RgR_g up from base. [B1]
  3. Friction FF at base towards wall. [B1]
  4. Normal reaction from wall RwR_w horizontal away from wall at top.
    (c) Take moments about base.
    Clockwise moment: Rw×(Lsin⁡60∘)R_w \times (L \sin 60^\circ). [M1]
    Anti-clockwise moment: W×(L2cos⁡60∘)W \times (\frac{L}{2} \cos 60^\circ). [M1]
    Equilibrium: RwLsin⁡60∘=WL2cos⁡60∘R_w L \sin 60^\circ = W \frac{L}{2} \cos 60^\circ [M1]
    Rw=Wcos⁡60∘2sin⁡60∘=W2tan⁡60∘R_w = \frac{W \cos 60^\circ}{2 \sin 60^\circ} = \frac{W}{2 \tan 60^\circ}
    Rw=W23R_w = \frac{W}{2\sqrt{3}} or 0.289W0.289 W. [A1]

15.
(a) Frictional force between tires and road. [B1]
(b) Centripetal force Fc=mv2rF_c = \frac{mv^2}{r}.
Max friction Ff=μR=μmgF_f = \mu R = \mu mg.
μmg=mv2r\mu mg = \frac{mv^2}{r} [M1]
v2=μgrv^2 = \mu gr
v=0.80×9.81×50v = \sqrt{0.80 \times 9.81 \times 50} [M1]
v=392.4=19.8 m s−1v = \sqrt{392.4} = 19.8 \text{ m s}^{-1}. [A1]
(c) Banking allows the normal reaction force to have a horizontal component. [B1]
This component helps provide the centripetal force, reducing reliance on friction. [B1]

16.
(a) vx=20cos⁡30∘=17.32 m s−1v_x = 20 \cos 30^\circ = 17.32 \text{ m s}^{-1}. [B1]
vy=20sin⁡30∘=10.0 m s−1v_y = 20 \sin 30^\circ = 10.0 \text{ m s}^{-1}. [B1]
(b) Time to max height: vy=uy+at⇒0=10−9.81t⇒t=1.02 sv_y = u_y + at \Rightarrow 0 = 10 - 9.81 t \Rightarrow t = 1.02 \text{ s}.
Total time =2×1.02=2.04 s= 2 \times 1.02 = 2.04 \text{ s}. [M1] [A1] (Allow 2.0 s if g=10g=10)
(c) Range =vx×t=17.32×2.04=35.3 m= v_x \times t = 17.32 \times 2.04 = 35.3 \text{ m}. [M1] [A1]

17.
(a) Force down slope =mgsin⁡30∘= mg \sin 30^\circ.
ma=mgsin⁡30∘⇒a=gsin⁡30∘ma = mg \sin 30^\circ \Rightarrow a = g \sin 30^\circ. [M1]
a=9.81×0.5=4.905 m s−2a = 9.81 \times 0.5 = 4.905 \text{ m s}^{-2}. [A1]
(b) Constant velocity means zero acceleration, so net force is zero.
Friction =Component of weight down slope= \text{Component of weight down slope}.
Ff=mgsin⁡30∘=2.0×9.81×0.5=9.81 NF_f = mg \sin 30^\circ = 2.0 \times 9.81 \times 0.5 = 9.81 \text{ N}. [M1] [A1]

18.
(a) Resultant R=102+102=200=14.1 NR = \sqrt{10^2 + 10^2} = \sqrt{200} = 14.1 \text{ N}. [M1] [A1]
(b) tan⁡θ=1010=1\tan \theta = \frac{10}{10} = 1.
θ=45∘\theta = 45^\circ. [M1] [A1]

19.
(a) Fnet=Thrust−Weight=15000−(1000×9.81)=5190 NF_{\text{net}} = \text{Thrust} - \text{Weight} = 15000 - (1000 \times 9.81) = 5190 \text{ N}. [M1]
a=Fnetm=51901000=5.19 m s−2a = \frac{F_{\text{net}}}{m} = \frac{5190}{1000} = 5.19 \text{ m s}^{-2}. [A1]
(b) As mass mm decreases, and Thrust is constant, the net force increases (since weight decreases). [B1]
Since a=F/ma = F/m, and FF increases while mm decreases, acceleration increases. [B1]

20.
(a) mgh=12mv2⇒v=2ghmgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}. [M1]
v=2×9.81×0.20=3.924=1.98 m s−1v = \sqrt{2 \times 9.81 \times 0.20} = \sqrt{3.924} = 1.98 \text{ m s}^{-1}. [A1]
(b) Air resistance is negligible / No energy loss to heat/sound. [B1]