A: Kinematics, forces, momentum, energy, rigid bodies and relativity – IB Physics exam questions with mark scheme

1Core4 marks

A car of mass 900 kg travels at a constant speed of $10\ \mathrm{m\,s^{-1}}$ around a circular roundabout of radius 25 m.

(a) Calculate the angular speed of the car. [1]

(b) Calculate the centripetal acceleration of the car. [1]

(d) State the direction of the resultant force. [1]

$\omega = v/r = 10/25 = 0.40\ \mathrm{rad\,s^{-1}}$ (0.40)(1)
$a = v^2/r = 4.0\ \mathrm{m\,s^{-2}}$ (allow $\omega^2 r$)(1)
$F = ma = 900 \times 4.0 = 3600\ \mathrm{N}$ (3600 N; allow $3.6\times10^3$ N)(1)
Towards the centre of the circle (allow: perpendicular to velocity / radially inward)(1)
Answer: (a) 0.40 rad s$^{-1}$ (b) 4.0 m s$^{-2}$ (c) 3600 N (d) towards the centre of the circle
Examiner tip: In (c) use the mass in kilograms and the centripetal acceleration; do not add a separate 'centrifugal force'.

2Standard5 marks

A stone is thrown horizontally at $15\ \mathrm{m\,s^{-1}}$ from the top of a cliff 45 m above the sea. Ignore air resistance and use $g = 9.81\ \mathrm{m\,s^{-2}}$.

(a) Determine the time taken for the stone to reach the sea. [2]

(b) Calculate the horizontal distance from the foot of the cliff at which the stone lands. [1]

Vertical motion from rest: $s = \tfrac12 g t^2$ (initial vertical velocity is zero)(1)
$t = \sqrt{2\times45/9.81} = 3.03\ \mathrm{s}$ (3.0 s)(1)
$x = u t = 15\times3.03 = 45\ \mathrm{m}$ (45 m; allow 45.4 m)(1)
Vertical speed $v_y = gt = 29.7\ \mathrm{m\,s^{-1}}$ (29.7)(1)
Speed $=\sqrt{15^2+29.7^2} = 33\ \mathrm{m\,s^{-1}}$ (33)(1)
Answer: (a) 3.03 s (b) 45 m (c) 33 m s$^{-1}$
Examiner tip: Treat horizontal and vertical motion separately and combine the components as vectors at the end.

3Standard5 marks

A block of mass 5.0 kg slides down a rough plane inclined at $30^\circ$ to the horizontal. The coefficient of dynamic friction between the block and the plane is 0.20. Use $g=9.81\ \mathrm{m\,s^{-2}}$.

(a) Calculate the component of the weight of the block parallel to the plane. [1]

(b) Determine the frictional force acting on the block. [2]

$mg\sin30^\circ = 5.0\times9.81\times\sin30^\circ = 24.5\ \mathrm{N}$ (24.5 N)(1)
Normal reaction $R = mg\cos30^\circ = 42.5\ \mathrm{N}$ (42.5 N)(1)
$F = \mu R = 0.20\times42.5 = 8.5\ \mathrm{N}$ (8.5 N)(1)
Resultant force down the plane $= 24.5-8.5 = 16.0\ \mathrm{N}$ (16 N)(1)
$a = F/m = 3.2\ \mathrm{m\,s^{-2}}$ (3.2)(1)
Answer: (a) 24.5 N (b) 8.5 N (c) 3.2 m s$^{-2}$
Examiner tip: Friction depends on the normal reaction, which is $mg\cos\theta$ on a slope, not $mg$.

4Standard5 marks

A trolley of mass 0.80 kg moves at $1.5\ \mathrm{m\,s^{-1}}$ along a frictionless track and collides with a stationary trolley of mass 0.40 kg. The trolleys stick together.

(a) Determine the common speed of the trolleys immediately after the collision. [2]

(b) Calculate the kinetic energy lost in the collision. [2]

Total momentum is conserved: $0.80\times1.5 = (0.80+0.40)v$(1)
$v = 1.0\ \mathrm{m\,s^{-1}}$ (1.0)(1)
Initial KE $=\tfrac12(0.80)(1.5)^2 = 0.90$ J and final KE $=\tfrac12(1.2)(1.0)^2=0.60$ J(1)
Loss $= 0.30\ \mathrm{J}$ (0.30 J)(1)
Inelastic, because kinetic energy is not conserved (some KE transferred to thermal/sound energy)(1)
Answer: (a) 1.0 m s$^{-1}$ (b) 0.30 J (c) inelastic
Examiner tip: Quote the conservation principle in words or symbols; a bare answer for speed can lose the first mark.

5Standard4 marks

A footballer kicks a stationary ball of mass 0.16 kg. The graph shows how the force exerted by the foot on the ball varies with time while they are in contact.

(a) Determine the impulse given to the ball. [2]

(b) Calculate the speed of the ball as it leaves the foot. [1]

Impulse equals the area under the force–time graph (e.g. two triangles plus a rectangle)(1)
Impulse $= 0.4+0.8+0.4 = 1.6\ \mathrm{N\,s}$ (1.6)(1)
$v = \Delta p/m = 1.6/0.16 = 10\ \mathrm{m\,s^{-1}}$ (10)(1)
$F_{av} = J/\Delta t = 1.6/0.06 = 27\ \mathrm{N}$ (27 N; allow 26.7)(1)
Answer: (a) 1.6 N s (b) 10 m s$^{-1}$ (c) 27 N
Examiner tip: Work out areas from the axes values, and remember a ball starting at rest has zero initial momentum.

6Standard4 marks

A motor raises a lift of total mass 800 kg through a vertical height of 12 m at constant speed in 15 s. The overall efficiency of the motor and lift system is 75%. Use $g=9.81\ \mathrm{m\,s^{-2}}$.

(a) Calculate the gain in gravitational potential energy of the lift. [1]

(b) Calculate the useful output power. [1]

$\Delta E_p = mg\Delta h = 800\times9.81\times12 = 9.42\times10^4$ J ($9.4\times10^4$ J)(1)
$P_{out} = 94176/15 = 6278\ \mathrm{W}$ ($6.3\times10^3$ W)(1)
Efficiency = useful power output / total power input, so $P_{in}=P_{out}/0.75$(1)
$P_{in} = 8371\ \mathrm{W}$ ($8.4\times10^3$ W)(1)
Answer: (a) $9.4\times10^4$ J (b) $6.3\times10^3$ W (c) $8.4\times10^3$ W
Examiner tip: Divide, not multiply, by the efficiency when finding input power.

7StandardHL only6 marks

A flywheel of moment of inertia $0.45\ \mathrm{kg\,m^2}$ is initially at rest. A constant torque of $3.0\ \mathrm{N\,m}$ is applied for 5.0 s. Friction is negligible.

(a) Calculate the angular acceleration of the flywheel. [1]

(b) Calculate its angular speed after 5.0 s. [1]

(d) The work done by a constant torque is $W=\tau\theta$, where $\theta$ is the angle turned through. Show that the work done by the torque during the 5.0 s is consistent with your answer to (c). [2]

$\alpha=\tau/I = 3.0/0.45 = 6.7\ \mathrm{rad\,s^{-2}}$ (allow 6.67)(1)
$\omega = \omega_i+\alpha t = 0+6.67\times5.0 = 33\ \mathrm{rad\,s^{-1}}$ (allow 33.3 or 33.5 from rounded $\alpha=6.7$; allow ECF from (a))(1)
$E_K=\tfrac12 I\omega^2$ with correct substitution(1)
$E_K = 250\ \mathrm{J}$ (allow 245–252 J from rounded values; allow ECF from (b))(1)
Angle turned $\theta=\omega_i t+\tfrac12\alpha t^2 = \tfrac12\times6.67\times5.0^2 = 83$ rad (allow $\theta=\tfrac12(\omega_i+\omega_f)t$ or $\omega_f^2=2\alpha\theta$; allow 83–84 rad)(1)
Work $=\tau\theta = 3.0\times83.3 = 250$ J, equal to $E_K$ in (c) (allow 249–252 J from rounded values) and statement that they agree because, with negligible friction, all the work done by the torque becomes rotational KE (OWTTE)(1)
Answer: (a) 6.7 rad s$^{-2}$ (b) 33 rad s$^{-1}$ (c) 250 J (d) $W=\tau\theta=250$ J
Examiner tip: Use unrounded values from earlier parts in (c) and (d) so the two energies agree.

8HardHL only6 marks

Cosmic-ray muons are created high in the atmosphere and travel towards the Earth's surface at speed $0.995c$. The proper lifetime of a muon is $2.2\ \mu\mathrm{s}$. Take $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$.

(a) Calculate the Lorentz factor $\gamma$ for the muon. [2]

(b) Determine the average lifetime of the muon as measured in the Earth frame. [1]

(d) Determine the distance, measured in the muon's frame of reference, that the Earth's atmosphere moves past the muon during its lifetime. [2]

$\gamma = 1/\sqrt{1-v^2/c^2} = 1/\sqrt{1-0.995^2}$(1)
$\gamma=10.0$ (10.0)(1)
$\Delta t=\gamma\Delta t_0 = 22\ \mu\mathrm{s}$ (22 μs)(1)
$d=vt=0.995\times3.00\times10^8\times22.0\times10^{-6} = 6.6$ km (6.6 km)(1)
Length contraction in the muon frame: $L=L_0/\gamma$(1)
$L = 6575/10.01 = 657$ m (about 660 m; allow $0.995c\times2.2\ \mu$s)(1)
Answer: (a) 10.0 (b) 22 μs (c) 6.6 km (d) about 660 m
Examiner tip: Be clear which frame each quantity is measured in: proper time is in the muon frame, proper length is in the Earth frame.

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