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 11, 2016

2007: Problems 28-30

Topic: Rigid BodiesConcepts: Fictitious forces, Statics Solution: We go to the accelerating frame to reduce the setup to a statics problem, introducing the fictitious inertial force −Ma→-M\vec{a} on the bicycle’s CM. Choosing the CM as the pivot point, we can balance torques: Since f1=μN1f_1=\mu N_1 and f2=μN2f_2=\mu N_2, Balancing forces in the vertical direction, N1+N2=MgN_1+N_2=Mg
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July 11, 2016

2007: Problem 27

Topic: Rigid BodiesConcepts: Artificial gravity, Conservation of angular momentum Solution: Let ω\omega be the initial angular velocity of the space station. In the rotating frame, all objects experience a (fictitious) centrifugal force of magnitude mac=mω2rma_c=m\omega^2r which looks like gravity mgmg if we have After two astronauts climb into the center, the space station rotates at
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July 11, 2016

2007: Problem 26

Topic: DynamicsConcepts: 5 kinematics equations, Inclined plane, Newton’s laws Solution: Applying Newton’s 2nd law to the sled, Since f=μN=μmgcos⁡θf=\mu N=\mu mg\cos\theta, If the sled travels at acceleration aa for time tt starting from rest, the distance it covers is Solving for μ\mu, so the answer is B.
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July 10, 2016

2007: Problem 25

Topic: Oscillatory MotionConcepts: Buoyant force, Simple harmonic motion Solution: Starting at equilibrium, suppose we push the pogo down by a distance xx. Then the buoyant force increases by the weight of the water displaced: Applying Newton’s 2nd law to the pogo (taking downwards to be positive), we have By definition of acceleration, a=x¨a=\ddot{x} so This
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July 10, 2016

2007: Problem 24

Topic: KinematicsConcepts: Projectile motion, Vectors Solution: There is a constant net force FnetF_{\text{net}} acting on the ball so it has to be thrown opposite to FnetF_{\text{net}} to come back to its starting position. We have so the answer is B.
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July 10, 2016

2007: Problem 23

Topic: GravityConcepts: Circular motion, Statics Solution: For the person at the North Pole who is at rest, we have For the person at the equator who is undergoing uniform circular motion, we have Thus, we have so the answer is C.
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July 10, 2016

2007: Problem 22

Topic: EnergyConcepts: Kinetic energy, Work-energy theorem Solution: Since a constant force FF acts on the rocket over a distance dd, the work done on the rocket is W=FdW=Fd. By the work-energy theorem, this is equal to the kinetic energy gained by the rocket, If the same force FF acts over the same distance dd but
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July 10, 2016

2007: Problem 21

Topic: Rigid BodiesConcepts: Moments of inertia Solution: Recall the moment of inertia of a sphere is Since M∝ρR3M \propto\rho R^3, If the density ρ\rho is the same and the radius RR is doubled, the moment of inertia II increases by a factor of 25=322^5=32 so the answer is E.
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July 10, 2016

2007: Problem 20

Topic: CollisionsConcepts: Kinetic energy Solution: The kinetic energy is given by Since each collision reduces EE by a factor of 22, vv is reduced by a factor of 2\sqrt{2}. We have so 6 collisions would reduce vv by a factor of 88. Thus, the answer is C.
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July 10, 2016

2007: Problem 19

Topic: EnergyConcepts: Conservation of energy, Spring potential energy Solution: We have a non-Hookian spring with force F(x)=−kx2F(x)=-kx^2 so its corresponding potential energy is Once the mass is released, it will momentarily be at rest when the loss in gravitational potential energy balances the gain in spring potential energy: Since θ=30∘\theta=30^{\circ} and h=dsin⁡θ=d/2h=d\sin\theta=d/2, so the answer
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July 10, 2016

2007: Problem 18

Topic: EnergyConcepts: Conservation of energy, Energy dissipation, Inclined plane Solution: The initial gravitational potential energy of the ice gets dissipated by friction: Since θ=30∘\theta=30^{\circ}, so the answer is B.
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July 9, 2016

2007: Problem 17

Topic: KinematicsConcepts: Projectile motion, Vectors Solution: Since the object is undergoing projectile motion, its acceleration is always directed downward with magnitude gg. At any point along the trajectory, we can split gg into a tangential component parallel to the velocity and a perpendicular component. The direction of the velocity vector can be determined from its
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July 9, 2016

2007: Problem 16

Topic: CollisionsConcepts: Conservation of linear momentum Solution: By conservation of linear momentum (taking upward to be positive), Solving for vv, so the answer is B.
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July 9, 2016

2007: Problem 15

Topic: System of MassesConcepts: Pulleys with rotational inertia Solution: Applying Newton’s 2nd law to the mass, we have Applying Newton’s 2nd law to the disk, we have Since there is no slipping, a=Rαa=R\alpha: Adding this equation to the first equation, we obtain The tension is then so the answer is B.
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July 9, 2016

2007: Problem 14

Topic: DynamicsConcepts: Forces in mechanics Solution: Friction points in the direction to oppose relative motion between two surfaces. Since the rear wheel is being powered and turned clockwise such that its bottom point is moving backward, friction on the rear wheel points in the forward direction. Since the front wheel is being pushed forward, friction
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July 9, 2016

2007: Problem 13

Topic: CollisionsConcepts: 1D elastic collision Solution: Recall the elastic collision equations for mass mm colliding with mass MM: If m=Mm=M, then so the first particle goes to rest while the second particle picks up the initial velocity of the first. Thus, their momenta swap after the collision so the answer is D.
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July 9, 2016

2007: Problem 12

Topic: DynamicsConcepts: Statics, Torque from weight Solution: Balancing torques around the fulcrum, we have using the fact that we can put all the weight of the measuring stick at its center of mass to find the equivalent torque. Since l1=0.2ml_1=0.2\,\mathrm{m}, we have l2=0.5m−l1=0.3ml_2=0.5\,\mathrm{m}-l_1=0.3\,\mathrm{m}. Thus, so the answer is C.
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July 9, 2016

2007: Problem 11

Topic: Rigid BodiesConcepts: Angular kinematics, Kinetic energy, Moments of inertia Solution: Recall for rotational motion, From kinematics, Thus, the kinetic energy of the objects after time tt is given by The torques on all the objects are the same so Recall the moments of inertia: Hence, so the answer is E.
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July 9, 2016

2007: Problem 10

Topic: Rigid BodiesConcepts: Rotational Newton’s 2nd law Solution: Recall for rotational motion, We are given the moment of inertia of the wheel: The force is applied perpendicular to the moment arm: Thus, Both wheels have the same mass, so for both to have the same angular acceleration, Hence, so the answer is D.
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July 9, 2016

2007: Problem 9

Topic: System of MassesConcepts: Dynamics of CM Solution: There is the external force of gravity in the vertical direction so the center of mass moves vertically with increasing speed. Note that there are no external forces in the horizontal direction. Thus, the answer is D.
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