Answer key to final test
1: B work is W = F(dot)d = Fx cos dx = (50 N) (8 m) cos 45 = (400 J) cos 45 = 280 J
2: B W = F d = (mg) (h) = (1 kg)(10 m/s2)(8 m) = 80 J
3: B U = W = 80 J. [If number 2 was incorrect and number 3 was the same, that showed the correct reasoning and was therefore marked correct]
4: A Force needed to push up slope is less than (mg). Fx = (mg)(5/20) = mg/4 = 100 N
5: A U = mgh = K = mv2/2; v2=2gh = 2(10 m/s2)(5 m) = 100 m2/s2 ; v=10m/s
6: A K = mv2/2 = W = Fd =FNd = mgd ; = v2/(2gd) =(20)(20)/[(2)(10)(200)]= 0.1
7: C hammer transfers kinetic energy into work done on nail
8: B U = kx2/2 = K = mv2/2; v2=kx2/m = (200)(0.2)2/(0.1) = 200(10)/[(5)(5)] =
80 m2/s2 ; v=9 m/s
9: A K = mv2/2; no height dependence
10: C U = mgh, so 2h leads to U = mg(2h) = 2(mgh)
11: B Both quantities of work are 500 J, it doesn’t matter if the work increases K or U
12: B temperature is a measure of KE, KE of projectile lost to heat -> KE of air
13: B at top, forces are not equal and opposite, so acceleration is not zero
14: A see Q5. If now 2h, v2=2g(2h) ; v = 1.4 v0
15: D KE is converted to PE and back, total energy is constant
16: C Power = E/t ; E = W (work-energy theorem)
17: D In-class discussion
18: B II: incompressible, so flow rate is constant. III: uniform cross-section, so v is constant. From Bernoulli’s equation, h must be equal at both ends
19: D Pressure at top and bottom is 1 atm, v at top is 0; v2=2gh
20: A With plug in hole, pressure is higher at point 2 by gh (gauge pressure)
21: A From first paragraph, fluid “piles up” under this condition and so is compressible
22: B some energy is converted to heat during bounce
23: B 2D kinematics: motion in y-axis independent of motion in x-axis. Cannonball II reaches smaller height, both have a=-g along y-axis, so II must hit first.
24: C gM on Moon is GM/r2 . Moon is M/100 and R/4, so gM = g (42/100) = 0.16 g
25: B r=2R, so F = GMm/r2 = 1/4 value at r=R.
26: C v2 = GM/R = (6.67e-11)(6e24)/(6e6) = 6.7e7, so v~103 m/s
27: A For rock at exactly escape velocity, all KE is converted to PE: mv2/2 = GMm/R
28: D Newton’s Third Law
29: C sound wave diffract around corners
30: E (B and D) sound waves are compressional waves; wave must be traveling for you to hear words
31: C v = f ; = v/f = (343m/s)/(700 1/s)~0.5 m
32: A wavefronts are compressed relative to violinist at rest, so frequency is higher
33: C conservation of momentum: m1v1 + m2v2 = (500 kg m/s – 250 kg m/s) = +250 kg m/s = MV = (150 kg)V ; V = (250 kg m/s)/(150 kg m/s) = 1.7 m/s to left (+x)
34: C conservation of momentum MV = 0 = m1v1+m2v2= m1v1+(2 kg)(10 m/s) ; m1v1 = -20 kg m/s ; v1 = (-20 kg m/s)/(50 kg) = -0.4 m/s (opposite direction of ball)
35: B mAvA + mBvB = MV = 0. mAvA = mA(-10vB) = -mBvB ; mA = mB/10
36: B vx is same, vy is same, ay=-g is same, so penguins land in same spot
37: C vy is same, ay=-g is same, so t is same; ax = 0, so dx = vxt, so faster penguin goes twice the distance