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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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Topic: DynamicsConcepts: Limiting cases, Statics Solution: We consider three times t1<t2<t3t_1<t_2<t_3 during the rotation. Time t1t_1 is the original setup where we have by balancing forces. Time t2t_2 is when the tensions swap, Time t3t_3 is the final setup where we have since it can’t have a horizontal component T2x=0T_{2x}=0 which means We now go
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Topic: DynamicsConcepts: Forces in mechanics, Limiting cases, Newton’s laws Solution: Since the box is moving at constant velocity, we balance forces in the vertical direction and horizontal direction: When the block is far away, θ→0\theta \rightarrow 0 so we have When the block is close to the pulley, θ→π/2\theta \rightarrow \pi/2 so we have Since
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Topic: Rigid BodiesConcepts: Moments of inertia, Rolling down inclined plane Solution: Recall the acceleration of an object rolling without slipping down an inclined plane is given by where the object has moment of inertia Objects A, B, and D are all solid balls with β=2/5\beta=2/5 so they reach the bottom at the same time. Object
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Topic: GravityConcepts: Escape velocity, Gravitational force Solution: Setting mgmg equal to the gravitational force, Recall the escape velocity vescv_{\text{esc}} is given by Since gR=GM/RgR=GM/R, so the answer is C.
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Topic: EnergyConcepts: Conservation of energy, Energy dissipation, Spring-potential energy Solution: By conservation of energy, the initial kinetic energy of the block is converted to spring potential energy and energy dissipated by friction. We have Solving for xx using the quadratic equation, where we took the positive root since x>0x>0 under our sign convention. Thus, the
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Topic: KinematicsConcepts: Fictitious forces, Free fall Solution: We can go to the free fall frame accelerating downwards at gg so that the projectiles become free particles (since we include the fictitious inertial force of magnitude mgmg pointing upwards which cancels out gravity). Then we have so the answer is B. Topic: KinematicsConcepts: Free fall Solution:
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Topic: CollisionsConcepts: 1D inelastic collision, Kinetic energy Solution: The initial kinetic energy is given by Conserving linear momentum, we have where vfv_f is the final velocity of both masses. The final kinetic energy is given by The loss in kinetic energy is so the answer is D.
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Topic: System of MassesConcepts: Conservation of linear momentum Solution: We have conservation of linear momentum since there are no external forces. Initially, there is no momentum so the final momentum vectors of the particles should add to zero. Since two particles emerge at right angles, the final momenta form the sides of a right triangle:
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Topic: DynamicsConcepts: Inclined plane, Statics, Torque from weight Solution: When the triangle is about to topple over, the normal force and friction only act at the bottom right corner. If we choose this point as our pivot point, then only the weight WW of the triangle contributes to the torque. At the threshold of toppling,
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Topic: KinematicsConcepts: Projectile motion Solution: Recall the range and height equations from projectile motion: We want to find the critical angle when H=RH=R, Since sin(2θ)=2sinθcosθ\sin(2\theta)=2\sin\theta\cos\theta, so the answer is A.
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Topic: KinematicsConcepts: Free fall Solution: Let TT be the time between when drops are released from the faucet. We are given that a drop takes time 2T2T to fall to the sink so The drop in the air has fallen for time TT so it is located a distance dd below the faucet with Thus,
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Topic: Rigid BodiesConcepts: Inclined plane, Rolling down inclined plane Solution: Recall the acceleration of an object rolling without slipping down an inclined plane is given by where the moment of inertia of the object is I=βmr2I=\beta mr^2. Recall the acceleration of a block sliding down an inclined plane is given by Since the hollow cylinder
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Topic: Rigid BodiesConcepts: Power, Torque Solution: The power supplied is given by where τ\tau is the torque supplied to the ring and ω\omega is the rotation rate. The torque supplied to the ring has to balance the torque from friction so Increasing RR increases the power required while increasing δ\delta has no effect at our
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Topic: GravityConcepts: Gauss’s law for gravity, Kepler’s laws, Projectile motion Solution: Thus, the answer is E.
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Topic: EnergyConcepts: Power Solution: Recall the instantaneous power is given by so lines of constant power are hyperbolas τ∝1/ω\tau \propto 1/\omega on a τ\tau–ω\omega plot. The furthest hyperbola (maximum power) is tangent to the graph at a point between II and III so the answer is D.
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Topic: OtherConcepts: Dimensional analysis Solution: We can find the form of n(Vm,T,R,σ)n(V_m, T, R, \sigma) via dimensional analysis. We have the units of all variables: Since [n]=mol[n]=\mathrm{mol}, we must have n∝R−1n \propto R^{-1}. Then so we have to divide by TT to cancel out K\mathrm{K} and obtain n∝R−1T−1n \propto R^{-1}T^{-1}. Then so we have to
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Topic: Oscillatory MotionConcepts: Mass-spring system, Vectors Solution: Since r→=xi→+yj→\vec{r}=x\vec{i}+y\vec{j}, we have F→=−kr→\vec{F}=-k\vec{r} so we can identify the spring constant as k=8N/mk=8\,\mathrm{N/m}. The period of a mass-spring system is given by The particle first returns to the origin in half the period, so the answer is C. Topic: EnergyConcepts: Conservation of energy, Spring potential energy Solution:
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Topic: EnergyConcepts: Energy dissipation, Inclined plane Solution: The loss in gravitational potential energy is due to the energy dissipated by friction. Thus, we have Simplifying, so the answer is A.
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Topic: DynamicsConcepts: Statics Solution: To keep the rope from slipping, the friction force must balance the weight of Becky: Since for static friction f≤μsNf \leq \mu_s N, so this is the minimum normal force Jonathan needs to exert. Therefore, the answer is E. Topic: EnergyConcepts: Energy dissipation Solution: From the previous problem, Jonathan applies a
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Topic: EnergyConcepts: Conservation of energy, Vertical spring Solution: The initial energy of the mass-spring system is where xx is the new equilibrium and dd is the subsequent displacement. Note we have the initial position of the mass as Ug=0U_g=0. When the mass passes through the new equilibrium, the energy is From energy conservation, Since kx=mgkx=mg,
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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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