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A-Level Physics – Mechanics

Kinematics, projectiles, forces, momentum, energy, circular motion, SHM.

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12 questions  ·  46 marks  ·  answers included  ·  A-Level Physics (Edexcel/AQA/OCR style)
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This worksheet covers:

  • Kinematics
  • Projectiles
  • Newton’s laws
  • Momentum
  • Work/energy
  • Circular motion
  • Simple harmonic motion

Calculator allowed; use $g=9.81\,\text{m s}^{-2}$ throughout. Allow about 50 minutes.

Questions

Section A – Kinematics and forces (skills)

1

A car accelerates uniformly from $5.0\,\text{m s}^{-1}$ to $25\,\text{m s}^{-1}$ in $8.0\,\text{s}$.(a) Calculate the acceleration of the car.(b) Calculate the distance travelled during this time.

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(a) $a=\dfrac{25-5.0}{8.0}=2.5\,\text{m s}^{-2}$.(b) $s=ut+\tfrac12at^2=5.0(8.0)+0.5(2.5)(8.0)^2=120\,\text{m}$.
[3]
2

A ball is dropped from rest from a height of 20 m above the ground (air resistance negligible).(a) Calculate the speed of the ball when it lands.(b) Calculate the speed of the ball after it has fallen 15 m.

Show answer
(a) $v=\sqrt{2gh}=\sqrt{2(9.81)(20)}=19.8\,\text{m s}^{-1}$.(b) $v=\sqrt{2(9.81)(15)}=17.2\,\text{m s}^{-1}$.
[3]
3

State Newton’s second law of motion, and use it to explain why a resultant force is needed to keep an object moving in a circle at constant speed.

Show answer
Newton’s second law: the resultant force on an object is proportional to its rate of change of momentum, $F=ma$, in the direction of the resultant force. In circular motion at constant speed the velocity direction is continuously changing, so there is a centripetal acceleration directed toward the centre; by $F=ma$ this requires a resultant (centripetal) force directed toward the centre, even though speed is constant.
[3]
4

A block of mass 5.0 kg rests on a frictionless plane inclined at $30^\circ$ to the horizontal.(a) Draw (describe in words) the forces acting on the block.(b) Calculate the acceleration of the block down the slope.

Show answer
(a) Weight $mg$ acting vertically downward; normal reaction $N$ acting perpendicular to the incline surface (no friction as the plane is frictionless).(b) $a=g\sin30^\circ=9.81\times0.5=4.9\,\text{m s}^{-2}$ (down the slope).
[3]

Section B – Exam-style: projectiles, momentum and energy

5

A ball is kicked with initial speed $20\,\text{m s}^{-1}$ at $30^\circ$ above the horizontal from level ground. Air resistance is negligible.(a) Calculate the horizontal and vertical components of the initial velocity.(b) Calculate the time of flight.(c) Calculate the horizontal range and the maximum height reached.

Show answer
(a) $v_x=20\cos30^\circ=17.3\,\text{m s}^{-1}$, $v_y=20\sin30^\circ=10.0\,\text{m s}^{-1}$.(b) $t=\dfrac{2v_y}{g}=\dfrac{2(10.0)}{9.81}=2.04\,\text{s}$.(c) Range $=v_xt=17.3\times2.04=35.3\,\text{m}$. Max height $=\dfrac{v_y^2}{2g}=\dfrac{100}{19.62}=5.10\,\text{m}$.
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6

A trolley of mass 2.0 kg moving at $6.0\,\text{m s}^{-1}$ collides with a stationary trolley of mass 4.0 kg. The trolleys stick together on impact.(a) Calculate the common velocity immediately after the collision.(b) Calculate the kinetic energy lost in the collision, and state what has happened to this energy.

Show answer
(a) By conservation of momentum: $v=\dfrac{(2.0)(6.0)+0}{2.0+4.0}=2.0\,\text{m s}^{-1}$.(b) $E_{k,\text{before}}=\tfrac12(2.0)(6.0)^2=36\,\text{J}$, $E_{k,\text{after}}=\tfrac12(6.0)(2.0)^2=12\,\text{J}$. Energy lost $=36-12=24\,\text{J}$; this energy is transferred to heat and sound (and to permanent deformation) during the (perfectly inelastic) collision.
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7

A block of mass 10 kg slides from rest down a rough slope of length 5.0 m, inclined at $20^\circ$ to the horizontal. The coefficient of friction between block and slope is 0.20.(a) Calculate the normal reaction force and the friction force on the block.(b) Calculate the speed of the block at the bottom of the slope.

Show answer
(a) $N=mg\cos20^\circ=10(9.81)\cos20^\circ=92.2\,\text{N}$. Friction $=\mu N=0.20\times92.2=18.4\,\text{N}$.(b) Resultant force down slope $=mg\sin20^\circ-\mu N=33.6-18.4=15.1\,\text{N}$, so $a=1.51\,\text{m s}^{-2}$. Using $v^2=u^2+2as=2(1.51)(5.0)=15.1$, $v=3.9\,\text{m s}^{-1}$.
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8

State the principle of conservation of momentum. A firework rocket of total mass 0.80 kg is moving at $30\,\text{m s}^{-1}$ when it explodes into two fragments. One fragment of mass 0.50 kg continues forward at $50\,\text{m s}^{-1}$. Calculate the velocity of the other fragment immediately after the explosion.

Show answer
Conservation of momentum: in an isolated system (no external resultant force) the total momentum before an event equals the total momentum after. Momentum before $=0.80\times30=24\,\text{kg m s}^{-1}$. Mass of second fragment $=0.80-0.50=0.30\,\text{kg}$. $24 = (0.50)(50)+(0.30)v \Rightarrow 24-25=0.30v \Rightarrow v=-3.3\,\text{m s}^{-1}$, i.e. $3.3\,\text{m s}^{-1}$ in the direction opposite to the rocket’s original motion.
[4]

Section C – Challenge: circular motion and SHM

9

A car of mass 900 kg travels around a flat, unbanked circular bend of radius 50 m at constant speed $20\,\text{m s}^{-1}$.(a) Calculate the centripetal force required.(b) Calculate the minimum coefficient of friction between the tyres and road needed for the car to complete the bend without skidding.

Show answer
(a) $F=\dfrac{mv^2}{r}=\dfrac{900(20)^2}{50}=7200\,\text{N}$.(b) Friction provides the centripetal force, so $\mu mg \geq \dfrac{mv^2}{r} \Rightarrow \mu \geq \dfrac{v^2}{rg}=\dfrac{400}{50(9.81)}=0.82$.
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10

A road is banked at angle $\theta$ so that a car can round a bend of radius 80 m at $25\,\text{m s}^{-1}$ with no reliance on friction. Calculate $\theta$.

Show answer
For no friction needed, $\tan\theta=\dfrac{v^2}{rg}=\dfrac{625}{80(9.81)}=0.796$, so $\theta=38.5^\circ$.
[3]
11

A mass of 0.50 kg is attached to a spring of spring constant $200\,\text{N m}^{-1}$ and undergoes simple harmonic motion with amplitude 0.050 m on a frictionless horizontal surface.(a) Calculate the angular frequency $\omega$ and the period $T$ of the oscillation.(b) Calculate the maximum speed and maximum acceleration of the mass.

Show answer
(a) $\omega=\sqrt{k/m}=\sqrt{200/0.50}=20\,\text{rad s}^{-1}$. $T=\dfrac{2\pi}{\omega}=0.314\,\text{s}$.(b) $v_{\text{max}}=\omega A=20\times0.050=1.0\,\text{m s}^{-1}$. $a_{\text{max}}=\omega^2A=400\times0.050=20\,\text{m s}^{-2}$.
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12

For the mass–spring system in the previous question, explain, with reference to the defining equation of SHM ($a=-\omega^2x$), how the acceleration and restoring force vary as the mass passes through the equilibrium position and through the extremes of its motion.

Show answer
At the equilibrium position $x=0$, so $a=-\omega^2x=0$: the restoring force and acceleration are zero and speed is maximum. At the extremes of the motion (displacement $=\pm A$), $|x|$ is maximum, so the magnitude of the acceleration (and restoring force) is maximum ($\omega^2A$), directed back toward the equilibrium position, while the speed is momentarily zero.
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