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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: EnergyConcepts: Work-energy theorem Solution: By the work-energy theorem, We have so the answer is A. Topic: EnergyConcepts: Kinetic energy Solution: From the previous problem, we found the initial kinetic energy is Solving for the mass, Then the initial momentum is so the answer is E.
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Topic: EnergyConcepts: Kinetic energy Solution: The kinetic energy is given by for the same momentum. Since the truck has ten times more mass M=10mM=10m, so the answer is D.
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Topic: DynamicsConcepts: Newton’s laws, Scale reading Solution: The scale reads the normal force NN and assumes it is the weight mgmg. The measured mass is then mscale=N/gm_{\text{scale}}=N/g. If the elevator has acceleration aa upward, we have so In the beginning, the elevator is at rest so a=0a=0 and mscale=m=80kgm_{\text{scale}}=m=80\,\mathrm{kg}. Once the elevator starts moving, the
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Topic: DynamicsConcepts: Statics, Torque from weight Solution: We choose the hinge as the pivot point and balance torques on one of the legs. The weight mgmg of the leg acts at its center while the normal force N=mgN=mg acts at the bottom to keep the leg at rest. Friction ff points inward to prevent the
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Topic: KinematicsConcepts: Limiting cases, Projectile motion Solution: Thus, the answer is E.
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Topic: KinematicsConcepts: Law of reflection, Projectile motion Solution: When the ball hits the wall, the horizontal component of the velocity flips direction. From that point onwards, the path taken is the reflection of the trajectory the ball would have taken without the wall. Thus, for the ball to reach Diego with the wall present, we
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Topic: KinematicsConcepts: 5 kinematics equations Solution: Let each car have length LL. If the train has acceleration aa, then it takes time t1t_1 where for the first car to pass the observer. Since t∝Lt \propto \sqrt{L}, the time t10t_{10} it takes the first 10 cars to pass the observer is Similarly, the time t9t_9 it
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Topic: GravityConcepts: Circular orbits, Conservation of angular momentum, Elliptical orbits Solution: Conserving angular momentum between the apogee and perigee of the ellipse, we have Since 2R≤r1≤3R2R \leq r_1 \leq 3R and R/3≤r2≤R/2R/3 \leq r_2 \leq R/2, we can bound Recall the velocity of a circular orbit of radius rr is given by We can relate
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Topic: DynamicsConcepts: Circular motion, Newton’s laws Solution: When the assembly is rotated, the spring forces on the mass provide the centripetal acceleration. Since the springs are stretched by distance ll, the spring forces have magnitude klkl. Applying Newton’s 2nd law, From the figure, we have rcosθ=lr\cos\theta=l so Since we have an equilateral triangle, θ=π/6\theta=\pi/6 and
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Topic: OtherConcepts: Dimensional analysis Solution: The dimensions of gg are so it is necessary (though not sufficient) to have equipment that can measure quantities with length dimensions and quantities with time dimensions. Thus, the answer is C.
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Topic: OtherConcepts: Dimensional analysis Solution: To figure out the base units of power, recall the equation P=FvP=Fv. Taking the units of both sides, To figure out the base units of force, recall the equation F=maF=ma. Taking the units of both sides, Substituting this in the previous equation, Since by definition [P]=W[P]=\mathrm{W}, we have Finally, so
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Topic: DynamicsConcepts: Effective spring constant, Vertical spring Solution: Note that the 10N10\,\mathrm{N} platform only shifts the equilibrium height of the spring system. We can neglect it when calculating how much an external force compresses the system, since the change in height is independent of the starting height. Thus, we have a system of three springs:
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Topic: FluidsConcepts: Buoyant force, Newton’s laws, Scale reading Solution: Note that the scales determine mass by dividing their measured force by gg. Thus, in the beginning, the first scale measures the normal force on itself as F1=M1gF_1=M_1g. The second scale measures its tension as F2=M2gF_2=M_2g. Once the block is submerged in the water, the buoyant
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Topic: FluidsConcepts: Newton’s laws, Power Solution: The motor has to supply a force FF on the water in the pipe that balances its weight WW since the water is moving at constant speed. Because the pipe is a cylinder, the mass of the water is Then The power required to pump the water at velocity
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Topic: Oscillatory MotionConcepts: Equivalence principle, Mass-spring system, Vertical spring Solution: Note that dropping the box changes the effective gravity experienced by the mass (in the box frame). Thus, the answer is B.
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Topic: OtherConcepts: Data expression Solution: From the graph, we have a line with slope (7−3)/(2−0)=2(7-3)/(2-0)=2 and intercept 33. Thus, TT and AA are related by so the answer is A.
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Topic: Oscillatory MotionConcepts: Effective spring constant, Mass-spring system Solution: The two springs are connected in parallel since both are connected to the block on one end and the cart on the other end. We can move the second spring to the same side as the first to make this clear: Spring constants in parallel add
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Topic: EnergyConcepts: Inclined plane, Power Solution: To drive up with constant velocity vv, the car needs to be pushed with force FF balancing gravity, The power needed to travel at velocity vv is FvFv. If the engine provides power PP and we assume no dissipation, the maximum speed is so the answer is A.
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Topic: Rigid BodiesConcepts: Statics, Torque from weight Solution: By the principle of superposition, drilling a hole is equivalent to superimposing negative mass at the same location. Thus, we consider the original cylinder with a negative mass cylinder of weight W=80N−65N=15NW=80\,\mathrm{N}-65\,\mathrm{N}=15\,\mathrm{N} added to the same location that would have been drilled. We choose the contact point
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Topic: EnergyConcepts: Work, Work-energy theorem Solution: The cart initially starts with speed v0v_0 in the negative direction under the influence of a positive force, so negative work −W0-W_0 is done on it until it comes to rest. Then the cart reverses direction and has positive work W0W_0 done on it when it returns to the
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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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