Here are five problems designed to test students' ability to use fundamental physics equations with the requirement of algebraic transformations:
Problem: Projectile Motion A ball is launched from the ground at an angle θ with an initial velocity v0. It reaches a maximum height H.
- Given: H and θ
- Find: The initial velocity v0
- Formula Involved: H=2gv02sin2θ
Solution Steps:
- Start with the given formula: H=2gv02sin2θ
- Rearrange to solve for v0:v02=sin2θ2gHv0=sin2θ2gH
Problem: Kinetic and Potential Energy Conversion A block of mass m slides down a frictionless incline of height h starting from rest. Find the speed of the block at the bottom.
- Given: m, h
- Find: The speed v at the bottom
- Formula Involved: mgh=21mv2
Solution Steps:
- Start with the energy conservation equation: mgh=21mv2
- Cancel out m from both sides:gh=21v2v2=2ghv=2gh
Problem: Uniform Circular Motion An object is moving in a circle of radius r with a constant speed v. Find the centripetal force Fc acting on the object.
- Given: m, v, r
- Find: The centripetal force Fc
- Formula Involved: Fc=rmv2
Solution Steps:
- Use the centripetal force formula: Fc=rmv2
- This problem is straightforward and requires just one algebraic step, but students can be asked to first derive v2 if given T (the period):v=T2πrv2=(T2πr)2=T24π2r2Fc=m⋅rT24π2r2=T24π2mr
Problem: Newton’s Second Law with Friction A block of mass m is pushed across a horizontal surface with a constant force F. The coefficient of kinetic friction between the block and the surface is μk. Find the acceleration a of the block.
- Given: F, m, μk
- Find: The acceleration a
- Formula Involved: F−Ffriction=ma and Ffriction=μkmg
Solution Steps:
- Start with the force balance: F−μkmg=ma
- Rearrange to solve for a:a=mF−μkmga=mF−μkg
Problem: Harmonic Oscillator A mass m attached to a spring with spring constant k oscillates with an amplitude A. Find the maximum speed vmax of the mass.
- Given: m, k, A
- Find: The maximum speed vmax
- Formula Involved: Etotal=21kA2=21mvmax2
Solution Steps:
- Start with the energy conservation formula: 21kA2=21mvmax2
- Cancel out the 21 terms:kA2=mvmax2vmax=Amk
These problems are designed to make students use their algebraic skills to manipulate the given equations before arriving at the final answer.