2nd Year Physics – 1

Unit 2 Test Review Answers:
1. b 2. b 3. a 4. e 5. b
6. c 7. b 8. d 9. e 10. c
11. T
1
cos 32 = T2
cos 45 so T1
= 0.834T2
Also, T1 sin 32 + T2
sin 45 = 100 , so
substitution
gives (0.834T
2) sin 32 + T2 sin 45 = 100
Therefore, T2 = 87N and T1
= 72.6N
Fs
3(9.8) sin35
12. FN
= = = 84.3N , so Fapp = FN
− Fgy
= 60.2N
μs
.2
13. The acceleration of the plane (and ring) is found by a = 65 − 0 = 2.17m/s 2 .
30
To calculate the angle of the string, think about the two components of the
tension force in the string. The x-component is causing the acceleration, T x =
ma. The y-component is balancing out the weight of the ring, so T y = mg. So
θ = tan -1 (ma/mg) = 12.5 o
14. On the table, T=9a. For the block hanging over the side, 29.4-T=3a.
2
Substitution gives a = 2.45m / s so T = 22.05N. Then the block’s
2 2
final speed is found with v = v + 2ax to be v = 2.43m/s.
o
15. First convert 78rpm to 1.225m/s. Then find the centripetal acceleration by
2
v
ac = = 10.0m/s 2 . The direction of the acceleration is toward the center
r
of the circular path.
16. For the car not to skid, friction must cause the centripetal acceleration, so
2
v
Ffric
= m(ac
). Ffric
=μ FN
= 13,132N. So the equation Ffric
= m yields r
v = 14.03m/s.
17. For the car to stay on the track, gravity must provide the force causing the
centripetal acceleration (with F N =0 to imply that the car is almost ready to
2
v
leave the track). So, Fg
= m yields a speed of 14 m/s.
r
18. Gravity must be the cause of the centripetal acceleration, F g = m , but F g is
found with Newton’s Universal Law of Gravitation. This yields the expression
GmEarth
v = which gives an orbital speed of 1019m/s. The period is
8
3.84 × 10
d 2πr
found with the equation v = =
6
which gives T= 2.37 × 10 s , or approx.
t T
27.4 days.
a c