Download Physics Practice Exam 2 Solutions

Physics Practice Exam 2 Solutions
1. C If p=mv and KE= ½ mv², and we want to put these two together so that way we are
left with v. You see that if we divided KE by p, m would cancel out, and one v, and we
would be left with ½ v, so v= 2KE/p. If we put the values from the table, you will see
that C is the correct answer.
2. A In inelastic collisions, some energy is lost due to compressing, or sticking together,
there fore A is false.
3. D The block’s kinetic energy is going to be dissipated by the friction on the ramp, and
because of work done by gravity. So you get an equation 38J= Wg + Wf = mgh +
μkmgdcosθ = mgdsinθ + μkmgdcosθ. h=dsinθ because of the trigonometry of the triangle.
Solving for d, you have d= 38/(4•9.8•sin30+ 4•9.8•0.72•cos30)= 0.86 m
4. B At the top, the ball has a potential energy of mgh, and that is equal to the energy
stored in the spring at the initial position, wince we have no outside forces. Therefore,
mgh=½kx², solve for k, and you get k= mgh/(½x²)=(0.8)(9.8)(3.4-0.04)/(½(0.04)²= 33
kN/m.
5. A Mechanical Energy is conserved in this problem, and at x=4, the particle has a
potential energy of 20J, and if its velocity 3 m/s, its kinetic energy is ½mv²= ½(2)(3)²=9J.
So, the place closest to the origin it will get in looking at the graph is about 2m, because
that is where the potential energy is 29, and it will move back away from the origin when
it gets there.
6. E The point of highest kinetic energy will be the point of lowest potential energy,
which is when U= -60J. Since mechanical energy is constant at all times, The kinetic
energy at this point must be 89 J, because The total mechanical energy, as found in
problem 5, is 29J. So, if 89J = ½ mv², then v= 9.43 m/s.
7. E The work done will be the change is potential energy, which is equal to mg∆y. We
can treat the cylinder as a point mass at its center of mass, which lying on its side, is 0.5m
off the floor. If I tip it up on its side, its center of mass is going to be ½ its height, which
is 1m, so its final center of mass is 0.5m off the floor. So mg∆y=(30)(9.8)(0.5-0.5)=0.
8. A Angular momentum is conserved, which is L=Iω. So, initially, I= ½ mR². The
final I= ( ½ mR² + ½ (2m) (R/3)². So I0ω0=Ifωf , so ω at the end =9/11 ω, after
rearranging and dividing.
9. D Taking the sum of torque around the hinge on the board, you get 0= - (√(7²-4²) 1.2) (40)(9.8) + F (4/7) √(7²-4²) – (25)(9.8)(√(7²-4²)/2) . The 4/7 comes from sin θ=
Opposite/Hypotenuse= 4/7. Solving for F, you should get 757 N.
10. E Angular momentum is conserved, which is L=mrv=Iω, because our only initial
angular momentum is from the bullet, then it is the bullet and the rod. Rearranging, v=
Iω./(mr)=((1/12 (4)(1)² + (0.003)(0.2)² ) (0.80) / ((0.003)(0.2) = 445 m/s .
11. C T=2π/ω=2π/(√(g/l)=2π/√(9.8/2)=2.84 s
12. B Using conservation of energy, U=mg(L-Lcosθ)=(5)(9.8)(2-2cos50)= 35 J. We use
50 degrees, because, looking at the trig of the triangle, you need the height above the
lowest position. So 35= ½ mv², solving for v, we should get 3.74 m/s.
13. D Using conservation of energy, mgh + ½ mv² (initial)= mgh + ½ mv² (final), the
mass will drop out, solving for final v, we get v= √(2( ½ (25)² + (9.8)(1.5) – (9.8)(12) )=
20.5 m/s
14. B Using sum of torque about the hinge, Iα=(55)(2.2), solving for α, and using I= 1/3
mL², you should get 3.34 rad/s².
15. E Calculating the critical b value, b=√(2•km)=√(2•100•0.400)=12.64. This
eliminates all choices except B and E, which we know B can’t be right because we know
the system moves. Also, note, kinetic energy cannot be negative.
** Note: The choices for 15 may be different than what was in class—the correct
answer looks like a decay curve (B or E, depending on when you saw it) **
16. C Using impulse-momentum theorem, J=∆p=F∆t, so F= ∆p/∆t=mv/∆t. Using
conservation of energy, mgh= ½ mv², we find v= 86.4 m/s. So F= (1.5)(86.4)/(0.2)=
648N.
17. A L= r x p, using definition of cross product of equation sheet, L=(3-2) i + (-4-1) j +
(1-6) k= i-3j-5k.
18. D T=2π/ω=2π/√(k/m), solving for k, we get k=4π²m/T²= 8.77 N/m.
19. C Using ω’ on equation sheet, ω0²=k/m, b=F/v, so putting numbers into equation, we
get 1.51 rad/s.
20. E Using sum of torque about scale 1, you get 0= (-120)(1.7) + F2(4.0), F2= 51 N.
120= F1+F2, so F2= 69 N
21. B The projectile will have just enough energy to reach infinity, so KE + U=0,
½ mv²=GMm/r, and v=√(2GM/r), so plugging in our values, VQ/ VP=√( RP/RQ)=√(1/3),
so 1:√3