PHYS 150 TEST #3 ; 11/14/07 ; Dr. Holmes ;
NAME:
DO ALL SEVEN PROBLEMS. THE WORTH OF EACH
PART OF EACH PROBLEM IS MARKED NEXT TO THE SLOT FOR THE ANSWER. SHOW YOUR WORK
FOR PARTIAL CREDIT. ALL ANSWERS SHOULD BE IN MKS UNITS UNLESS OTHERWISE
INDICATED. Mearth = 6.0 x 1024 kg; Rearth
= 6,390 km.
1) The mass of the object in this problem is 55 kg. In all parts of this problem, neglect air resistance.
For parts a and b of this problem, assume that PE = mgh is valid.
a) What is the
amount of kinetic energy the object initially on the earth's surface [h=0])
needs to reach a height of 3,210 km
(2,000 miles)? [This assumes the object
will then simply fall back down.]
1.73 x 109 J.
b) What is the velocity of the object initially (at h=0 m.) with this assumption that PE = mgh?
7,932 m/s.
c) Using the more general form for gravitational potential energy, should your answer to the above two questions be bigger, the same, or smaller than what you get assuming PE = mgh ?
Smaller.
d) What would be the needed initial velocity of the object at the earth's surface to reach 3,210 km high (and then fall back down) if you use the more general form for gravitational potential energy?
6,472 m/s.
2) a) Design a dart gun (that is, specify the spring constant, k; the mass of the dart, m; and the distance that the dart is to be compressed when loaded, x, that will shoot the dart out with an exit speed (from the gun) of 8 m/s.
b) How high will this dart go if fired straight up?
3.26 m.
c) ) If you change the design so that the dart comes out twice as fast as in part a, will the dart go: [less than twice as high, twice as high, or more than twice as high] ?
More than twice as high.
d) In this part, indicate all answers that would work (e.g., if 1 and 2 are correct, write 1,2 as your answer; if none work, then write none. To double the exit speed of the dart, you could:
(1) double the spring constant;
(2) reduce the mass by half;
(3) double the distance you compress the dart.
3 .
3) A skier of mass 70 kg stands at the top of a hill that is 210 meters in vertical height above the valley below. The hill has a constant grade of 48o with the horizontal. In all parts neglect air resistance. (This is not a realistic assumption!)
a) Assuming no friction (m=0) and no attempt to slow down, and assuming the person starts from rest, how fast will the person be going when he/she reaches the bottom of the hill?
64.16 m/s = 144 mph .
b) If the grade of the hill were changed to 24 degrees (made less steep), would the speed at the base of the hill be less, the same, or greater (again assuming m =0) ?
Same.
c) If the person had an initial speed at the top of the hill of 4 m/s (again assuming m»0), would the final speed be greater [by an amount greater than 4 m/s, by an amount equal to 4 m/s, or by an amount less than 4 m/s]?
By an amount less than 4 m/s.
d) If the person started from rest at the top of the hill, but there was some friction present (say m=0.1), would changing the slope of the hill from 48o to 24o reduce the final speed, not change the final speed, or increase the final speed at the base of the slide?
Reduce the final speed.
e) Explain your answer to part d:
4) A car of mass 2,100 kg goes from rest to 30 m/s (65 mph) in 9 seconds. Neglect the energy lost to friction and air resistance in your answers below.
a) If the car started from rest, what is the car's final kinetic energy?
945,000 J.
b) What is the average power provided the car during the98 seconds in Watts?
105,000 Watts.
c) What is the average power provided the car in horsepower?
141 hp.
d) If the force provided by the engine was constant, was the power delivered by the engine [decreasing with increasing speed; constant; or increasing with increasing speed]?
Increasing with increasing speed.
5) For the following use the information in the plot of potential energy (PE) versus distance (x) shown below and the fact that the total energy of the object is +4 Joules, its mass is 0.25 kg, and there is no air resistance or friction.
a) What
is the particle's potential energy at x = 2.5 m?
-5 Joules.
b) What is the particle's kinetic energy at x = 2.5 m?
9 Joules.
c) What is the particle's speed at x = 2.5 m?
8.49 m/s.
d) Which direction is the force n the particle at x = 2.5 m? [left, right, none-zero, undeterminable with info given]
Left.
e) If the particle moves from the 2.5 m position towards the 3.0 m position, will its speed: [increase, stay the same, decrease, undeterminable with info given] ?
Decrease.
6) An object of mass, m1 = 1,500 kg, is moving South with a speed, v1i = 29 m/s, and hits the rear of another object of mass, m2 = 2,350 kg that is also moving South with a slower speed of v2i = 14 m/s.
a) Which object (#1 or #2) has the bigger initial kinetic energy?
#1.
b) Which object (#1 or #2) has the bigger initial momentum (magnitude) ?
#1.
c) If
the two masses stick together after the collision, and if all the external
forces can be neglected, then what will the speed of the combined masses after
the collision be ?
19.84 m/s.
d) What is the direction of the combined mass moving after the collision?
South.
e) Was momentum conserved during this collision?
Yes.
f) Was there any kinetic energy "lost" in this collision?
Yes.
7) A space vehicle of mass 6,000 kg (including crew) when empty of fuel takes on 44,000 kg of fuel at a space station. It is released from the station with an initial speed of 2 m/s, and its rockets throw out the gas at a speed of 5,200 m/s when turned on.
a) Can the rocket reach a speed of 5,200 m/s?
Yes.
b) If the answer is Yes, then how much fuel will it take to reach this speed? If the answer is No, then what is the minimum speed the fuel needs to be ejected at for the rocket to be able to reach a speed of 5,200 m/s ?
31,599 kg.
c) What is the maximum speed that the rocket can reach with the given amount of fuel and the given payload weight?
11,027 m/s