Transcript
PHYS 195 Spring 2018 Solution to Review Problems for Exam 1
50728492225812194477175563The first rock is thrown up with speed v: The second rock is thrown down with speed v:
The key is to find out what rock 1’s speed is when it passes down at the starting point. We can find this using v2 = vo2 + 2a(x-xo), and use the fact that when rock 1 passes on the way down, we want to find its speed at x = xo. We also use vo = v, per the given info above.
Thus we simply get v2 = v2, or v = +/- v. Since rock 1 is on the way down we use –v. This is the same velocity that the second rock has when it is thrown at the same height. Thus, both rocks will hit the ground with the same speed.
Rower 1’s velocity will angle downstream, after we add rower 1’s velocity and the steam velocity. Rower 2’s velocity would have the row velocity as the hypotenuse of the right triangle, to make sure the net velocity is straight across the stream.
Rower 2 only gets a portion of his velocity going straight across the stream (the other portion is rowing upstream to offset the stream’s velocity. Thus, rower 2 would take longer to reach the other side.
For a projectile that has velocity v, it has a component velocity of vxi + vyJ. vx is a constant. vy starts out positive, and decreases because of acceleration due to gravity. Thus, the magnitude of the velocity (the speed) would be lowest when vy = 0. This corresponds to the highest point in its path.
We use V1 = Vx1i + Vy1j + Vz1k and V2 = Vx2i + Vy2j + Vz2k. Then looking at just the i-comp for V1 X V2, we get: (Vy1Vz2 – Vz1Vy2). Now we reverse the signs: V1 = -Vx1i - Vy1j - Vz1k and V2 = -Vx2i - Vy2j - Vz2k. Again we look at just the i-comp for V1 X V2 to get: [(-Vy1)(-Vz2) – (-Vz1)(-Vy2)] = (Vy1Vz2 – Vz1Vy2), which is the same result as before. We can do the same technique for the j and k components.
We find the amount of water by first calculating the volume = length X width X height: volume = (5 km X 105 cm/km) X (8 km X 105 cm/km) X (1.0 cm) = 4 X 1011 cm3 then we use the conversion factor (in reality the density) of water to get the mass: mass = volume X density = (4 X 1011 cm3) X (10-3 kg/cm3) X (1 ton/103 kg) = 4 X 105 tons
Find the time for the sound to travel the length of the lane from: tsound = d/vsound = (16.5 m)/(340 m/s) = 0.0485 s. Find the speed of the ball from v = d/(T – tsound). Why do we subtract the times? Because the total time T is 2.50 seconds, the total of the ball moving down the lane and the sound moving back up the lane. Thus the travel time for the ball down the lane is the difference. Then v = (16.5 m)/(2.50 s – 0.0485 s) = 6.73 m/s.
+ Find the average acceleration from v2 = vo2 + 2a(x-xo); where v = 0. Then, 0 = [(95 km/h) X (1000 m/km) X (1 h/3600 s)]2 + 2a(0.80 m), which gives a = -4.4 X 102 m/s2. The number of g’s is: |a| = (4.4 X 102 m/s2)/[(9.80 m/s2)/g] = 44g.
The magnitudes of the vectors are: A = (Ax2 + Ay2 + Az2)1/2 = [(6.8)2 + (4.6)2 + (6.2)2]1/2 = 10.3 and B = (Bx2 + By2 + Bz2)1/2 = [(8.2)2 + (2.3)2 + (-7.0)2]1/2 = 11.0. Use the scalar product to find the angle: A B = AxBx + AyBy + AzBz = AB cos ?. So, (6.8)(8.2) + (4.6)(2.3) + (6.2)(-7.0) = (10.3)(11.0) cos ?. Then cos ? = 0.203, so ? = 78o.
(a) A X B = (AyBz – AzBy)i + (AzBx – AxBz)j + (AxBy – AyBx)k = [(-3.5)(2.0) – (0)(7.0)]i + [(0)(-8.5) – (7.0)(2.0)]j + [(7.0)(7.0) – (-3.5)(-8.5)]k = -7.0i – 14.0j + 19.3k (b) Magnitude of A X B = AB sin ?. A = [(7.0)2 + (-3.5)2]1/2 = 7.83, B = [(-8.5)2 + (7.0)2 + (2.0)2]1/2 = 11.2, and magnitude A X B = [(-7.0)2 + (-14.0)2 + (19.3)2]1/2 = 24.8, so 24.8 = (7.83)(11.2) sin ?, which gives sin ? = 0.283, so ? = 164o.
(a) Because the athlete lands at the same level, we can use the expression for the horizontal range: R = vo2 sin(2?o)/g and then 7.80 m = vo2 sin[2(33.0o)]/(9.80 m/s2), which gives vo = 9.15 m/s. (b) For an increase of 5%, the initial speed becomes vo’ = (1 + 0.05)vo = (1.05)vo, and the new range is: R’ = vo’2 sin(2?o)/g = (1.05)2vo2 sin(2?o)/g = 1.10R. Thus the increase in the length of the jump is: R’ – R = (1.10 – 1)R = (0.10)(7.80 m) = 0.78 m.
