Download Physics Practice Exam Solutions

Physics Practice Exam Solutions
February 14th
1. [A] Magnitude of A- B = √[(Ax-Bx)² + (Ay-By)²]=√[(-2.9-3.4)² + (7.0-(-4.2))²]= 12.85
2. [B] Magnitude of vector = √[(-4)²+(5)²+(10)²] = √141
Now find x, y, and z components in ratios to the magnitude, which gives you the
unit vector, so i= -4/√141=-0.33, j=5/√141=0.42, k=10/√141=0.84
3. [B] If you draw a free-body diagram, you have: sum of F=ma, if she weighs 1000N,
then her mass is 1000N/(9.8m/s²) =102.04 kg, and her acceleration is 4.9 m/s², so
F=ma=(102.04 kg) • (4.9 m/s²) = 500 N
4. [D] Using conservation of energy, we come up with the equation: 0.5kx²=mgh.
Solving for k, we get k=2mgh/x²=2(.180 kg)(9.8 m/s²)(1 m) / (0.20)² = 88.2 N/m
5. [D] In drawing a diagram of where John walks, you will find he is 1 km east from
where he started, so this is his total displacement. The total time is (3 km)/(2.5 km/hr) +
(2 km)/(15.0 km/hr) + (20 min) • (1 hr) / (60 min) + (30 min) • (1 hr) /(60 min) = 2.16667
hr. So the average velocity is (1 km)/( 2.16667 hr) = 0.46 km/hr
6. [E] If we have v(t), we can find a(t) just by taking a derivative, so a(t)= 3at² + 4at³ =
3t² + 8t³, so a(3)= 3(3)² + 8(3)³ = 243
7. [D] If you draw a straight line from City A to City B, this will be the sum of your
two vectors, the plane and the wind. If you put the plane and wind tip to tail with the sum
being the line between the cities, then you have a right triangle and you solve for the
cities vector: √(210²-150²)= 146.97 ( km/hr). Since the cities are 600 km apart, then
600 km / (146.97 km/hr) = 4.08 hr
8. [E] Studying the forces when the block stops moving on the incline, the maximum
static friction force is fs=μsN=(0.6)(1)(9.8)cos(30)=5.097 N, and the gravity pulling the
block down the ramp is mgsinθ=(1)(9.8)sin(30)= 4.905 N. The static friction force is
stronger than the gravity pulling the block down the ramo, so the block will never return
to the bottom of the ramp.
9. [B] Sum of F= msysasys=15N, solving for asys: asys=15N/(50 kg) = 0.3 m/s²
10. [C] We can find the F on each block, FA= (30 kg)(0.3 m/s²)= 9 N, and FB=
(20 kg)(0.3 m/s²)= 6 N. FA/ FB=9 N / 6 N = 1.5
11. [D] Taking along the string as a positive axis, the sum of F= msysasys= 3mg. Now we
can solve for asys: asys=(3mg)/(6m) = g/2
12. [C] You can draw a free body diagram at the top of the hill, and there is a gravity
and then the circular force as the car is moving in circular motion, in a cross-section
sense. Sum of F=0=mv²/r - mg, solving for v, we get v=√(gr)=√[(9.8 m/s²)(1024 m)]=
100.18 m/s
13. [C] Just as in Paula’s Lecture 10, you need to find the net force of the system: 60 N
– 40 N =20 N. Now find the acceleration of the system: asys=Fnet/msys=(20 N)/(10 kg) = 2
m/s². The sum of the forces on the string is: ma=FR-T, where m is the mass of the blocks
pulling on the string in the direction of motion, so T=FR-ma=60-4(2)=44N
14. [B] We need to find when static friction force equals the magnitude of the force
keeping the ball in circular motion, so when ma= μsmg, so when a= μsg, then μs=a/g.
There are two components of a, a radial and a tangential. So arad=rω² and mrω² 3.04 m/s²,
so μs=3.04/9.8=0.31.
15. [B] We can find the displacement, θ, in radians, by the equation ωf²-ω0²=2αθ, which
is a simple conversion from the 1-dimensional equation to circular motion. Solving for
θ=(4.00)²/(2 • 0.02)= 40 rad. Now, we just make a conversion from rad to revolutions: 40
rad • [(1 rev)/(2π rad)] = 6.4 rev
16. [A] This is a conservation of energy problem, setting it up, we get:
mgh+0.5mv²=0.5kx², because there is both potential and kinetic energy initially, then it is
all elastic potential at the end. Solving for x, we get:
x=√[(mgh+0.5mv²)/(0.5k)]=√[((5)(9.8)(2)+(0.5)(5)(10)²)/((0.5)(1000))]=0.834 m
Or, by using work-kinetic energy, the same numbers are also found.
17. [D] Using conservation of energy, in order to avoid some error to rounding, we can
use the total energy of the initial system: mgh2=mgh1+0.5mv², solving for h2, we get
h2=[(gh1+0.5v²)/g]=[(9.8•2 + 0.5•10²)/9.8]=7.10 m
18. [C] There are a few ways to do this problem. The way I would do it is using
conservation of energy: mgh + 0.5mv0²=0.5mvf², solving for h: h=[(0.5vf²0.5v0²)/(g)]=[(0.5(92)²-0.5(45)²)/9.8]=330 m
You can use projectile motion equations, also. We can find the y component if velocity
and solve for the height. Since vx=45cos30=38.97, and is constant for a projectile, the
final magnitude of velocity is 92, so 922=vx2+vy2, vy=√(922-38.972) = 83.34. Using
vf2=v02+2ad, where d is h, then h=(83.342-(45sin30)2)/(2*9.8)=328m.
19. [D] Using W=F•d=Fdcosθ, we have W=250•4•cos36=809 N.
20. [E] First find the sum of the two forces, which make a right angle: Sum of F =
√(FJ²+FS²)=√(100²+144²)= 175 N. The block is 1 kg, and a=F/m=175/10= 1.75 m/s²