Kevin S. Huang

October 26, 2016

F=ma Exam Solutions

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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July 14, 2016

2014: Problem 7

Topic: DynamicsConcepts: Statics, Torque from weight Solution: We balance torques around the pivot point. Using the fact that the weight of the stick can be taken to act at its center, we have Solving for mstickm_{\text{stick}}, so the answer is C.
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July 14, 2016

2014: Problem 6

Topic: CollisionsConcepts: Center of mass, Conservation of linear momentum, Dynamics of CM Solution: We will approximate the block’s position xx with the position xCMx_{\text{CM}} of the CM. Since both the block and the CM is within the cube of mass MM, we know To find xCMx_{\text{CM}}, note that there is no net external force on
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July 14, 2016

2014: Problem 5

Topic: DynamicsConcepts: Circular motion, Newton’s laws, Statics Solution: Applying Newton’s 2nd law, we have in the vertical direction Since friction provides the centripetal acceleration, We choose the CM as the pivot point and balance torques. The weight of the unicyclist acts at the CM so doesn’t contribute to the torque. If the unicyclist has length
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July 14, 2016

2014: Problem 4

Topic: System of MassesConcepts: Conservation of energy, Conservation of linear momentum Solution: 1st statement: If a particle at rest explodes into 2 particles, by conserving momentum and energy we obtain The first equation tells us the magnitudes of the momenta are equal p1=p2≡pp_1=p_2 \equiv p so which can be solved for pp and then we
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July 14, 2016

2014: Problem 3

Topic: FluidsConcepts: Buoyant force, Floating Solution: Recall the fraction of an object submerged in water is given by In our case, When the external force F=3NF=3\,\mathrm{N} acts on the object, it becomes fully submerged. Balancing forces on the object, Solving for the volume VV, Since ρobj=0.8ρwater\rho_{\text{obj}}=0.8\rho_{\text{water}}, so the answer is D.
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July 14, 2016

2014: Problem 2

Topic: Rigid BodiesConcepts: Inclined plane, Newton’s laws Solution: Since the ball is rolling down the incline, there is a normal force NN on the ball perpendicular to the incline and a friction force ff on the ball up along the incline. By Newton’s 3rd law, the ball exerts the same forces in the opposite directions
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July 14, 2016

2014: Problem 1

Topic: DynamicsConcepts: Angular momentum, Circular motion Solution: In the driver’s perspective, the position vector r→\vec{r} points left from the center of the circle to the car. The velocity vector v→\vec{v} always points forward. The angular momentum vector is given by By the right-hand rule, r→×v→\vec{r} \times \vec{v} points into the page (downwards for the driver)
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July 14, 2016

2013: Problem 25

Topic: DynamicsConcepts: Inclined plane, Newton’s laws Solution: Since the block slides down at constant speed with gravity and friction alone, we have Because N=Wcos⁡θN=W\cos\theta and f=μNf=\mu N, Thus, When a horizontal force PP is applied to the box, it moves up the ramp at constant speed so the net force on the box is still
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July 14, 2016

2013: Problems 23-24

Topic: EnergyConcepts: Conservation of energy, Forces in mechanics Solution: By conservation of energy, the gravitational potential energy of the man is converted into spring potential energy at the bottom of his fall: This is also when the tension in the spring is maximized so Substituting this into the first equation, Thus, so the answer is
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July 14, 2016

2013: Problem 22

Topic: DynamicsConcepts: Statics Solution: Choosing the ankle joint as the pivot point, only the normal force from the ground and tension in the Achilles tendon contribute to the torque: Since the student is standing on a single toe, N=mgN=mg so we have so the answer is D.
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July 14, 2016

2013: Problems 19-21

Topic: DynamicsConcepts: Circular motion, Newton’s laws Solution: Since the pendulum is undergoing circular motion, we have so the tension is At the bottom θ=0\theta=0, cos⁡θ\cos\theta is maximized since cos⁡0=1\cos\,0=1 while vv is also maximized by conservation of energy (pendulum is at lowest point of swing). Thus, both terms are maximized so the tension is largest
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July 14, 2016

2013: Problem 18

Topic: System of MassesConcepts: Conservation of angular momentum, Conservation of linear momentum, Dynamics of CM Solution: Since there are no external forces (and torques), we have conservation of linear momentum and angular momentum. The initial linear momentum is Since ptot=MtotvCMp_{\text{tot}}=M_{\text{tot}}v_{\text{CM}}, this means vCM=0v_{\text{CM}}=0 so the CM stays at rest at its initial location (1,0)(1, 0).
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July 14, 2016

2013: Problem 17

Topic: DynamicsConcepts: Fictitious forces, Gravitational force, Statics, Types of equilibrium Solution: We go to the rotating frame where the rod is at rest and include the (fictitious) centrifugal force directed outwards on all objects. Then, the external forces on our system are gravity from the planet and the centrifugal force. The gravitational force is given
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July 14, 2016

2013: Problem 16

Topic: OtherConcepts: Dimensional analysis, Gravitational force Solution: We can use dimensional analysis to find how the time TT of collapse scales with ρ0\rho_0, r0r_0, and GG: We first find the dimensions of each quantity: To find the dimensions of GG, we use so We take the dimensions of the first equation, Counting powers of time
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July 14, 2016

2013: Problem 15

Topic: FluidsConcepts: Buoyant force, Statics, Torque from weight Solution: We choose the hinge as our pivot point and balance torques on the rod. Only the weight of the rod and the buoyant force contribute to the torque. The weight WW acts downward at the CM of the rod, distance l/2l/2 away from the pivot. The
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July 14, 2016

2013: Problem 14

Topic: CollisionsConcepts: 1D elastic collision, CM frame Solution: Initially, mass mm is moving at 12m/s12\,\mathrm{m/s} towards the 4kg4\,\mathrm{kg} mass at rest. We go to the CM frame (moving at vCMv_{\text{CM}}): Recall the total linear momentum ptot=MtotvCMp_{\text{tot}}=M_{\text{tot}}v_{\text{CM}} so ptot=0p_{\text{tot}}=0 in the CM frame (since the CM is at rest by definition). Then after an elastic collision,
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July 14, 2016

2013: Problem 13

Topic: GravityConcepts: Angular kinematics, Circular orbits Solution: If the ring is part of Saturn, then it rotates together with the planet i.e. has the same angular velocity ω\omega. From angular kinematics, If the ring is a satellite of Saturn, then it moves at the velocity of a circular orbit which is Thus, the answer is
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July 14, 2016

2013: Problem 12

Topic: Rigid BodiesConcepts: Moments of inertia, Rolling down inclined plane Solution: Recall the acceleration aa of an object rolling without slipping down an incline of angle θ\theta is where the object has moment of inertia I=βmr2I=\beta mr^2. We will find the effective β\beta for our composite object of a spherical shell containing a frictionless fluid.
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July 14, 2016

2013: Problem 11

Topic: System of MassesConcepts: Inclined plane, Newton’s laws Solution: Since the triangular block remains at rest, the net force on it is zero. In addition to its own weight, the block feels the normal forces from the two cubes and the ground. As the cubes are effectively sliding down an incline, the normal force on
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July 14, 2016

2013: Problem 10

Topic: Rigid BodiesConcepts: Floating, Gauss’s law for gravity, Gravitational force, Rolling down inclined plane Solution: Thus, the answer is B.
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