Physics 101 Fall 2003 Homework Set 5 Solutions Exercises: ---------- Chapter 7: ---------- 2: Note that this question, like ALL non-numerical homework questions, requires both an answer and an explanation. The work required to stop the truck is equal to the kinetic energy of the truck. Thus, the question is which truck has more kinetic energy? First, we need to know which truck is moving faster. Both have the same momentum p = mv but one truck is heavier. We can think of it as Truck 1: p = Mv (big mass, small velocity) Truck 2: p = mV (small mass, big velocity) We can solve for the kinetic energy in terms of the mass and velocity one of two ways: symbolically or numerically. Symbolically: KE = (1/2)mv^2 = (1/2) (mv) v = (1/2) p v Since both trucks have the same momentum and the lighter truck has more velocity, the lighter truck will have more KE. Numerically: Plug in some arbitrary values. Let's try p = 10 kg m/s, truck 1 mass m1 = 1 kg and truck 2 mass m2 = 2 kg. This gives the velocity of truck 1: v1 = p / m1 = 10 kg m/s / 1 kg = 10 m/s the velocity of truck 2: v2 = p / m2 = 10 kg m/s / 2 kg = 5 m/s Now we can calculate the KE of truck 1: KE1 = (1/2) m1 v1^2 = (1/2) 1 kg (10 m/s)^2 = 50 J and truck 2: KE2 = (1/2) m2 v2^2 = (1/2) 2 kg (5 m/s)^2 = 25 J Thus, although both trucks have the same momentum, the lighter one has more kinetic energy! 4: The longer the barrel, the greater the distance over which the force acts. Since work = force * distance, as the length of the barrel increases, the amount of work done increases. This increases the kinetic energy (and hence the velocity) of the emerging bullet. 8: p = mv. If you double the mass and don't change the speed, then p doubles. Similarly, KE = 1/2 m v^2. If you double the mass and don't change the speed, then KE doubles. 10: The damage done by an object is proportional to the energy it carries (see the section on 'Comparison of Kinetic Energy and Momentum' pp 115-118 and especially figure 7.19). Both have the same energy so both are equally safe to catch. 16: Work is done when the force acts, at least partially, in the direction of motion. On a level road, gravity is perpendicular to the direction of motion so no work is done. On a down hill, the direction of motion is at least partially downward (in the direction of the force) so some work is done. 18: This is the same confusing third law question we always ask. If kick a soccer ball, it will accelerate because there is a large net force on the soccer ball. The motion of the soccer ball is NOT affected by the fact that the soccer ball exerts an equal and opposite force on your foot. Similarly, the rope exerts a force on the crate in the direction of motion. Therefore it does work on the crate. The fact that the crate exerts a force on the rope (and hence does work on the rope) is completely irrelevant to what happens to the crate. Only forces exerted ON the crate affect the crate. Forces exerted BY the crate affect other objects. 20: Energy is conserved. If you start with 1000 J of PE plus 0 J of KE and end up with 900 J of KE and 0 J of PE, then the remaining 100 J must have been converted to thermal energy (or heat). 34: In the case of a car, there are two horizontal forces on it 1) the force of the engine (transmitted through the transmission and the tires) and 2) the force of friction plus air resistance. These two forces total to zero. The engine still does positive work on the car (since the force is in the direction of travel). After all, the engine is burning gasoline to keep the car moving in the presence of friction. Thus, work is being done on the car. Friction is also doing work on the car. In this case negative work, since the force and the direction of motion are opposite. The NET force is zero, but there are still forces exerted on the car. The NET work is zero, but there is still work done on the car by the engine and by friction. 36: There are two ways to solve this problem. The energy way is to calculate the total energy (potential plus kinetic) of each ball as it is thrown. If the two balls have the same total energy when they are thrown, then they will have the same kinetic energy just before they hit the ground (since the total energy will have converted to potential energy). The total energy is the sum of the kinetic plus potential energy. Both balls have the same potential energy (since they are thrown from the same height). Both balls have the same kinetic energy (since they have the same magnitude of velocity). Therefore both balls have the same total energy and both will have the same kinetic energy before striking the ground. The other method is to remember figure 3.8 on page 47. That shows that a ball thrown upward with a certian speed will have exactly the same speed when it returns to the point from which it was thrown. Thus, a ball thrown upward with speed v will have the same downward speed v a short time later when it returns to the same height. This means that the upward thrown ball and the downward thrown ball will have the exact same downward velocity at that height so they will hit the ground with the same speed. Either solution is acceptable. Problems: --------- Chapter 7: 2: work needed to stop the car = KE. Since KE = (1/2) mv^2, if we double the speed, then we quadruple the KE. If the speed is 3 times larger, then the KE is 9 times larger. (You can plug in numbers to try it. It will work with any mass.) Since the braking force is unchanged, in order to do 9 times more work, the force has to be exerted over 9 times more distance. Therefore you will need a distance of 9 * 15 m = 135 m to stop. [skip this next part if it looks like gibberish to you] In equations (since F [the braking force] and m [the mass of the car] are unchanged): F d1 = 1/2 m v1^2 and F d2 = 1/2 m v2^2 or d1 = 1/2 m v1^2 / F and d2 = 1/2 m v2^2 / F Now I can divide the last two equations (divide the left side of the first by the left side of the second and the same for the right sides) d1/d2 = v1^2 / v2^2 = (150 km/h)^2 / (50 km/h)^2 = 9 d1 = 9 * d2 = 135 m 6: We need to compare the total kinetic energy before (in some arbitrary units) with the total kinetic energy after (in the same arbitrary units): The kinetic energy before = KE1 + KE2 = 1/2 m (10)^2 + 1/2 m (0)^2 = 50 m + 0 = 50 m The kinetic energy after = KE1 + KE2 = 1/2 m (5)^2 + 1/2 m (5)^2 = 12.5 m + 12.5 m = 25 m Thus the kinetic energy has decreased by 50% (or by 1/2) from 50 m to 25 m. You can also do this by plugging in an arbitrary mass (eg: 1 kg) and choosing arbitrary units for the velocity (eg: 10 -> 10 m/s). Estimation: ----------- How many dumptrucks of hurricane debris were removed from Norfolk and VB streets after hurricane Isabel? We need to know three numbers to estimate this. The population of the two cities is about 1/2 million = 5*10^5 (the actual number is about 640,000) (any number between about 250,000 and 1,000,000 is OK). There are about 2 people per household so there are about 2.5*10^5 homes. Each household had about 1 tree down (well, it was definitely less than 10 per house and more than 0.1 per house so I'll go with one per house). One dumptruck could haul away one good-sized tree. This means that they needed 2.5*10^5 dumptruck loads. Rather a LOT! (Any number within about a factor of ten of this is at least as good as mine.) Note that I have ignored people living in apartments and trees that fell in public property. Those offset each other (and anyway, I don't know the answer well enough to bother with small corrections).