Physics Practice Exam – Final

Solutions

1. AA third law pair always involves two opposing forces, since it is asked what the third law pair is with the force by the book on the table, the correct choice is the force by the table on the book.

2. A To start, sum the forces, and you will find that a=gsinθ-μgcosθ=5.62 m/s. Since the object starts from rest, the one-dimensional motion equation with constant acceleration is d= ½at². Solve for t, t=√(2d/a), and d= 4.5/cosθ, using trig. Plug in, and t=1.58s.

3. C An unstable equilibrium point is where the force is equal to zero and the force is greater than zero to the right and less than zero to the left. The easiest way to think of it is picturing it as a potential energy graph. At such a point, the potential is less either way you go, making it an unstable equilibrium, so B and F are unstable equilibrium points.

4. C Sum forces, and a=2/3 gsin(30)=1/3 g. The tension is T=ma (of the lagging block) which then is T=(3m)(1/3g)=mg.

5. E Writing out an energy balance: ½ kx1²= ½ kx2² + ½ mv² + μkmgd. If you plug in what we know, you get 0.18=0.02+0.15v²+0.094. Solving for v, you get v=0.66 m/s.

6. D Setting up conservation of energy, ½ Iω1² + mgd = ½ Iω2², where d is the height drop of the center of mass, which is the distance from the axis to the center of mass. Solving for I, you should get I=mgd/( ½ ω2² - ½ ω1²)= 4.9 kg m².

7. A Impulse is ∆p=Using the graph, The area under the curve is ½ (0.3)(50) + (0.5)(50) = 32.5. Since the force opposes the motion, our momentum equation looks like this: ∆p=m2v2-m1v1, solving for v2, you get 0.18 m/s.

8. C The spheres’ detachment causes no extra torque on the rod, so its angular speed is not affected.

9. E The work done by the electric field is W= ½ m(vf²-v0²)= ½ (2)(4²-0²)=16 J. W=Fd, so F=W/d =16/1=16 N. E=F/q=16/0.003=5333.3 V/m.

10. C P=I²R, so we need to find the current in the system, which will be equal to the current through the 4Ω resistor. R of the system is R= 1/ (1/3 + 1/3) + 4 = 11/2. I=V/R = 40/11. Putting this into first equation, we get 52.9 W.

11. B Using equation above, C=ε0 A/d= C=ε0 (π(0.20)²)/0.004 = 2.78E-10, which is 27.8E-9, so 28 nF is the correct answer.

12. D J=I/A=qnv. We know that v=d/t, so t=d/v. We solve the first equation for v, v= I/(Aqn)=(2.2)/(0.00145²• 1.60E-19 • 9E28 ) = 0.000042664. t= 40/ 0.000042664 gives us time in s, divide this by 3600 to get 153 hr.

13. A Brightness is proportional to the amount of current through the bulb. If two bulbs are in an isolated series system, then their brightness is equal, so B2 = B1. B3 has the same current through it as both B2 and B1 combined, so B3>B2 = B1. Where B4 is, it has the sum of all the current, so it is the brightest, making B the correct answer.

14. C First, we need to find the current through the system. V=IR, R= 1/(1/R1+1/R2) +R3 = 73.33. I=V/R= 240/73.33 = 3.27. V across R3 is = (3.27)(60)= 196.36 V. The rest of the potential drop will happen at the two resistor system. So, VR1=240-196.36=43.64. Then, I=V/R=43.64/20= 2.18 A.

15. A From our equation sheet, U is proportional to C at constant V. C=ε0 A/d, therefore, if you move from d to3d, C becomes C/3, which in turn U become U/3.

16. B The near frictionless surface allows the momentum of the collision to be conserved, but when the puck hits the wall momentum is lost the wall, making B the right choice.

17. D The cart is moving in the –x direction so the v is negative, eliminating all choices except C and D. Since the cart slows down, its velocity gets closer to 0, so D matches this description.

18. A Setting up a free body diagram and using trigonometry, ΣFy=414cos63 + 506cos48 = 527 N.