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Please find my solutions to past F=ma Contest problems below. Past Exams Problems organized by topic Quarter-final Exams Problems organized by topic
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Solution: A Concepts:
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Solution: B Concepts:
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Solution: A Concepts:
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Solution: D Concepts:
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Topic: EnergyConcepts: Energy dissipation, Inclined plane Solution: Between the start and end, the gravitational potential energy of the block is dissipated by friction. We have Solving for h2h_2, so the answer is A.
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Topic: CollisionsConcepts: Conservation of linear momentum, Impulse-momentum theorem, Relative velocity Solution: Since there are no external forces, we have conservation of linear momentum. Initially, the astronaut is at rest so the initial momentum After the launcher is fired, Since pi=pfp_i=p_f, We are given the relative velocity v1+v2=vrv_1+v_2=v_r so Solving for v1v_1, Thus, the impulse delivered
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Topic: GravityConcepts: Circular orbits, Gravitational force, Kepler’s laws, Reduced mass Solution: To simplify calculating the angular momentum and period, we use the concept of reduced mass to turn the two-body problem of masses mm and MM interacting via gravity into the equivalent one-body problem of a particle orbiting a fixed source. Recall the reduced mass
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Topic: System of MassesConcepts: Atwood machine, Moments of inertia, Pulleys with rotational inertia Solution: We first find the acceleration in an Atwood machine where the pulley has moment of inertia II. Applying Newton’s 2nd law to each mass and the pulley, Assuming there is no slipping α=a/R\alpha=a/R, we have Adding this equation and the first
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Topic: Rigid BodiesConcepts: Angular kinematics, Artificial gravity, Rotational Newton’s 2nd law Solution: Setting the centripetal acceleration equal to gg, we have so the space station needs to be rotating at angular velocity By Newton’s 2nd law, Since we have a ring, I=MR2I=MR^2 so From kinematics, The rockets need to fired for time so the answer
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Topic: DynamicsConcepts: Air resistance, Newton’s laws Solution: Every small piece of cable feels the same friction force f→\vec{f} to the left and the same weight w→\vec{w} downward since they both depend only on the length of the piece. Thus, all the pieces of rope effectively have weight w→′=w→+f→\vec{w}’=\vec{w}+\vec{f} towards the bottom left direction. We know
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Topic: DynamicsConcepts: Atwood machine, Statics Solution: The tension from the two strings must balance the weight of the pulley. We have The tension from the three strings must balance the weight of the box: Adding these two equations, we obtain We could have directly obtained this by considering the external forces on the pulley-box system.
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Topic: GravityConcepts: Gauss’s law for gravity Solution: By the shell theorem, the gravitational field outside a uniform sphere is the same as that of a point at the center of the sphere with the same mass. Thus, Since the radius is fixed, we have g∝ρg \propto \rho so it remains to find how the density
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Topic: CollisionsConcepts: 1D elastic collision, 1D inelastic collision, Limiting cases Solution: We study the two limiting cases of a perfectly inelastic collision and an elastic collision which will provide us with the bounds that apply to any generic collision. For a perfectly inelastic collision, the masses stick together so For an elastic collision, recall Thus,
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Topic: EnergyConcepts: Inclined plane, Power Solution: To move at constant velocity, a force must be supplied on the car. To travel at speed vv, the power required is since there are no energy losses. Because θ=30∘\theta=30^{\circ}, so the answer is A.
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Topic: Rigid BodiesConcepts: Kinetic energy, Rotational Newton’s 2nd law Solution: By Newton’s 2nd law, we have so a=T/Ma=T/M and α=TR/I\alpha=TR/I. The kinetic energies are given by Then by canceling out common factors, Multiplying the numerator and denominator by MIMI, so the answer is D.
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Topic: DynamicsConcepts: Air resistance Solution: Since the helicopter quickly reaches terminal velocity vtv_t, from kinematics, the time TT of flight is given by Thus, α=1\alpha=1 so the answer is E. Topic: OtherConcepts: Dimensional analysis Solution: From the previous problem, we found T=khrβρδWωT=khr^{\beta}\rho^{\delta}W^{\omega}. Taking the dimensions of both sides, Using the fact that [ρ]=M/L3[\rho]=M/L^3 and [W]=ML/T2[W]=ML/T^2,
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Topic: EnergyConcepts: Conservation of energy, Kinetic energy, Spring potential energy Solution: By conservation of energy, we have which is a line with slope −1-1 and intercept EE. Since K=12mv2≥0K=\frac{1}{2}mv^2 \geq 0 and U=12kx2≥0U=\frac{1}{2}kx^2 \geq 0, the plot only exists in the first quadrant. Thus, the answer is E.
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Topic: KinematicsConcepts: Angular kinematics, Circular motion Solution: Recall the centripetal acceleration is given by Since the car starts at ω0\omega_0 and has angular acceleration α\alpha, Substituting into the previous equation, We solve for the time tt it takes the car to reach the critical centripetal acceleration ac=243m/s2a_c=243\,\mathrm{m/s^2}, so the answer is E.
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Topic: CollisionsConcepts: Impulse-momentum theorem Solution: Recall the area under a FF–tt plot corresponds to the impulse JJ. By the impulse-momentum theorem, Solving for the final velocity vfv_f, In our case, going from t=0st=0\,\mathrm{s} to 7s7\,\mathrm{s}, we have J=12.5kgm/sJ=12.5\,\mathrm{kg\,m/s} so Thus, the answer is C.
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Topic: Oscillatory MotionConcepts: Effective spring constant, Mass-spring system Solution: From the period formula for a mass-spring system, we have By cutting the spring in half, the spring constant gets doubled since adding two half-springs in parallel should result in the original spring, Oscillating on an inclined plane doesn’t affect the period since it is independent
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1D elastic collision 1D inelastic collision 5 kinematics equations Air resistance Angular kinematics Angular momentum Artificial gravity Atwood machine Average vs. instantaneous Ballistic pendulum Bernoulli's principle Buoyant force Center of mass Circular motion Circular orbits CM frame Conservation of angular momentum Conservation of energy Conservation of linear momentum Coriolis force Data expression Dimensional analysis Dynamics Dynamics of CM Effective spring constant Elliptical orbits Energy dissipation Equivalence principle Error propagation Escape velocity Fictitious forces Floating Fma: Collisions Fma: Dynamics Fma: Energy Fma: Fluids Fma: Gravity Fma: Kinematics Fma: Oscillatory Motion Fma: Other Fma: Rigid Bodies Fma: System of Masses Forces in mechanics Free fall Gauss's law for gravity General angular momentum General kinematics equations General kinetic energy General Newton's 2nd law Gravitational force Gravitational potential energy Impulse-momentum theorem Inclined plane Kepler's laws Kinematics Kinetic energy Law of reflection Limiting cases Mass-spring system Mechanics Moments of inertia Motion graphs Newton's laws Parallel-axis theorem Pascal's law Perpendicular-axis theorem Physical pendulum Potential energy graphs Power Pressure with depth Projectile motion Pulleys with rotational inertia Qtrfin: E&M Qtrfin: Mechanics Recursion for rotational inertia Reduced mass Relative velocity Rigid Bodies Rolling down inclined plane Rolling motion Rotational Newton's 2nd law Scale reading Simple harmonic motion Simple pendulum Small angle approximation Speed vs. velocity Spring potential energy Statics Surface tension Tensile strength Torque Torque from weight Types of equilibrium Vectors Velocity constraints Vertical spring Wave speed in string Work Work-energy theorem Young's modulus
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