Homework 2 Solutions

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	PROBLEM 1 A top-fuel dragster acclerates from a standing start to cover 1320 feet in 4.75 s. What is the average acceleration of teh dragster. How long will the dragster take to stop and how far will it travel if its brakes and drag chute combined can uniformly decelerate the car at a maximum of 1g (1g = 32 ft/s2. The first part of this question asks for the average acceleration of the dragster. We'll use equation 1 from the above list, d = 1/2at2. This equation is valid since we have a uniformly accelerating object from 0 over a set time. Plug and chug: d = .5*a*t2 a = 2d / t2 a = 2*1320feet / 4.752s2
1/2 point	a = 117 feet/s2
	The second half of the problem wants to know the time and distance for stopping at 1G. The first critical part of this problem is to obtain the initial velocity. You can treat the first and second halves of this problem COMPLETELY SEPARATELY. So. Now we have this new problem where there's a car travelling at velocity v stopping at 1G. The question is what that initial velocity is. We can get it from the acceleration we just calculated. The velocity at the beginning of the stop, and the end of the 1320 feet, can be found from equation 2, v = at: v = a*t v = 117feet/s2*4.75s v = 556 feet/s Now we determine t2, the time to stop. We can use the same equation, just remember that this time the a and t are completely different than from the first half: v = a*t t = v / a t = 556 feet/s / 32feet/s2
1/4 point	t = 17.4 seconds
	With this time we then plug back into equation 3 to get the total distance covered in the stop: d = 1/2at2 d = .5 * 32feet/s2 * 17.42s2
1/4 point	d = 4827 feet
	PROBLEM 2 A girl kicks a soccer ball so that it rolls off a cliff at 30 m/s in the horizontal direction. If the cliff is 100 meters high, how far away from the base of the cliff will the ball first hit the ground? The key to this one is to recognize that this problem can be broken up into 2 separate problems, one in the horizontal and one in the vertical direction: Since the vertical part is the one that determines how long the fall lasts, we'll use it to figure out the total time (from equation 3): Note that in this instance the acceleration a is equal to the acceleration of gravity, g = 9.8 m/s2 dy = 1/2gt2 t = squareroot of 2d/a t = (2d/a)1/2 t = (2 * 100m / 10m/s2)1/2 t = 201/2 t = 4.5 seconds Now we use this time to figure out the distance travelled in the x or horizontal direction, along with equation 1: dx = vt dx = 30m/s * 4.5s
1 point	dx = 135m
	PROBLEM 3 A newly discovered Kuiper Belt object is found to have an orbit whose semi-major axis is 100 astronomical units. How long will it take the object to make one complete orbit of the sun? This problem is a direct application of Kepler's 3rd Law relating the period of a planet (time for 1 orbit) to the semimajor axis (distance from the Sun): P2 = r3 P2 = (100AU)3 P2 = 106
1 point	P = 1000 years
	PROBLEM 4 A soccer ball collides with an automobile window. If the soccer ball comes to a stop in 0.1 s before it bounces, and was traveling 40 m/s before the collision, what is the magnitude of the soccer ball's deceleration? If the soccer ball has a mass of 500 grams, what force is exerted by the car's windshield on teh ball (and by the ball on the windshield) in the process of stopping and reversing the ball's motion? The first part of this problem is to find the acceleration. Acceleration is normally a vector, having an amount AND a direction, so by asking only for the magnitude of the deceleration only the amount is required. We can use equation 2 here: v = at a = v/t a = 40m/s / .1s
1/2 point	a = 400m/s2
	Lastly we are asked to find the force exerted. According to equation 5 from Newton's Laws: F = ma F = 500g * 400m/s2 F = 200000 g*m/s2 Force is usually given in Newtons, however, one Newton being equal to 1 kg*m/s2. So we need to convert the g to kg: F = 200 kg*m/s2
1/2 point	F = 200 N
	PROBLEM 5 The Moon moves in a nearly circular orbit around theEarth of radius of 3.84*108m in 27.3 days. What is the centripetal acceleration of the Moon in m/s2? For this we will be using the equation for centripetal acceleration, #6. But in order to do so, we'll need to come up with the velocity. To do that we'll use the first equation, distance equals velocity times time: d = vt v = d/t But then we have to find the distance. The circumference of a circle is equal to 2*pi*r. v = 2*pi*r/t v = 2 * 3.14159 * 3.84*108m / 27.3 days v = 8.8*107 m/day An answer in meters per day isn't terribly helpful, lets convert that to meters per second: v = 8.8*107m day hour day 24 hours 3600 s v = 1000 m/s Now we use this velocity in the centripital acceleration equation: a = v2/r a = (1000m/s)2 / 3.84*108m
1 point	a = 2.7 * 10-3 m/s2
	Woah, this is an awefully small number! Does that make sense? Well, gravity exerts a 10m/s force here at Earth's surface, and the moon is 60 times as far from Earth as the surface, so by the inverse square law it should get 1/602 less gravity = 1/3600*10 = 1/360 = .00278. Very close, so the answer checks out. PROBLEM 6 What pairs of forces act in the following situations? a. A pitcher throws a fastball. b. An apple falls to the ground. c. The Moon orbits the Earth. d. A pencil rests on your desk. All I'm looking for here are forces that are acting in each situation, there might be 2, or 1, or more, but points were given for either the 1 present or 2 of what are there.
1/4 point each	a. Gravity earth on ball, gravity ball on earth, electrical forces from atoms in the ball and atoms in the pitcher's hand (can be called normal force) (both ways), drag (air resistance) (air on ball and ball on air heating it up). b. Gravity (both ways), drag (both ways), normal force or electrical forces between atoms when the apple hits the ground (both ways). c. Gravity. We know that the force of gravity is a centripetal force in this case, but both gravity AND centripetal force acting doesn't make any sense. So the answer is just gravity, in both directions. d. Gravity, electrical forces between atoms / normal force (both work both ways).
	PROBLEM 7 You are driving down Speedway Boulevard in a car when a pedestrian darts in front of you. If you have 10 meters in which to stop before striking the pedestrian, you and your car have a total mass of 1500 kg, and you are travelling at 20m/s, what is the minimum force that must be exerted to stop the car before it hits the pedestrian? Well, this problem asks for a force. There are several ways to do this problem. Here is the hard way: From equation 5 F=ma. So knowing the mass already we'll have to determine the acceleration before we can find the force. We'll get the acceleration by plugging a distance of 10m into equation 3, plugging in from equation 2 to eliminate acceleration in favor of time: v = at d = 1/2at2 d = 1/2vt t = 2d/v t = 20m / 20m/s t = 1s Now we plug this time back into the equation 2 to obtain the acceleration, and that into equation 5 to get the force. v = at a = v/t a = 20 m/s / 1s a = 20 m/s2 F = ma F = 1500kg * 20m/s2 F = 30000 kg*m/s2
1 point	F = 30000 N
	Here's an easier way using the concepts of energy that you used in Campus Cop: The amount of work done on the car must be equal to its initial kinetic energy if the car is brought to a complete halt: W = Fd KE = 1/2mv2 W = KE Fd = 1/2mv2 F = 1/2mv2/d F = .5 * 1500kg * (20m/s)2 / 10m F = 30000N
	PROBLEM 8 Tracie (50kg) and Tommie (75kg) are standing at rest in the center of the roller rink, facing each other, free to move. Tracy pushes off Tommie with her hands and remains in contact with Tommie's hands, applying a constant force for 0.75s. Tracy moves 0.5m during this time. When she stops pushing off Tommie, she moves at a constant speed. a. What is Tracy's constant acceleration during her time of contact with Tommie? b. What is Tracy's final speed after this contact? c. What force is applied to Tracy during this time? What is its origin? d. What happens to Tommie? If Tommie moves describe his motion, its force, acceleration, or Tommie's final velocity. a. Since Tracy is undergoing constant acceleration we can plug into equation 3: d = 1/2at2 a = 2d/t2 a = 2 * .5m / .752s2
1/4 point	a = 1.777 m/s2
	b. Using the acceleration from part (a) use equation 2 to find the final speed: v = at v = 1.777m/s2 * .75s
1/4 point	v = 1.333 m/s
	c. The force Tracie feels is a result of Newtons law that for every action there is an equal and opposite reaction; as she pushes on Tommie he pushes back on her. Use equation 5 with the acceleration from (a): F = ma F = 50kg * 1.777m/s2
1/4 point	F = 89N
1/4 point	d. Tommie receives the same force that Tracy did, 89N, from Newton's law. He moves away from her and from the point he was originally occupying. To find his acceleration, plug back into equation 5 with his mass. F = ma a = F/m a = 89 kg*m/s2 / 75 kg a = 1.33 m/s2 To get the final speed: v = at v = 1.33 m/s2 * .75s v = .89 m/s You can check this by using conservation of momentum (I had to) 50kg * 1.333m/s = 67 kg m/s 75kg * .89 m/s = 67 kg m/s Yup, works!
	PROBLEM 9 A truck with a mass of 20,000 kg and traveling at 50m/s collides with a 2000kg car that is just leaving a stoplight (the driver has released the brakes but hasn't started to accelerate). If the truck is moving straight ahead at 10 m/s immediately after the collision, how fast is the car moving immediately after the collision. If the collision lasts for .5s (that is, the time the truck and the car are in contact), how much force was exerted by the truck on the car. Draw a diagram of at least 2 action-reaction force pairs during the collision. This is a conservation of momentum problem, so the way to solve it is to figure out the momentum before the collision and after the collision (equation 9), set them equal, and solve for the final velocity. In the diagram above Vb is velocity of the truck before, Va is velocity after, and Vcb and Vca are the corresponding velocities for the car. pb truck = Mtruck * Vb pb car = Mcar * Vcb pb car = 0 (because Vcb is 0) pb total = pb truck + pb car pb total = pb truck = Mtruck * Vb pa truck = Mtruck * Va pa car = Mcar * Vca pa total = pa truck + pa car pa total = Mtruck * Va + Mcar * Vca Now set momentum before equal to momentum after for the conservation part: pb total = pa total Mtruck*Vb = Mtruck*Va + Mcar*Vca Mtruck*Vb - Mtruck*Va = Mcar*Vca Mtruck(Vb-Va) = Mcar*Vca (Vb-Va)(Mtruck/Mcar) = Vca Vca = (50m/s-10m/s)(20000kg/2000kg) Vca = (40m/s)(10)
<1/2 point>	Vca = 400m/s
	The second part of the question asks what the force on the car was if the collision lasted .5 seconds. We'll use Newton's trusty F=ma, but first we have to find the acceleration (equation 2): v = a*t a = v/t a = 400m/s / .5s a = 800m/s2 BE VERY CAREFUL HERE: You need to use the acceleration and mass for the SAME body, though it doesn't matter which body you use since the forces are equal and opposite. You could use the change in velocity of the truck, 40m/s, and its mass, 20000kg, and get the same answer. F = ma F = 2000kg * 800m/s2
1/4 point	F = 1.6 * 106 N
1/4 point for diagram
	Some people tried to solve this problem through conservation of energy instead of conservation of momentum. While normally this would work if the collision were elastic (if they bounced off one another instead of sticking together), the number 10m/s after the collision given in the problem as a speed for the truck is not sound physics. As it turns out the slowest the truck could be going after the collision is 42m/s if the collision were perfectly elastic, in which case the car's velocity would be 138m/s. Full credit was given for using the conservation of energy method.
	PROBLEM 10 What is the force required to keep Jupiter in orbit around the Sun, assuming that Jupiter move sin a circular orbit at 5 Astronomical Units from the sun (an Astronomical Unit is 1.5 * 108 km -- its usually just abbreviated to AU) and has a mass of 1.9 * 1027kg (about 318 times the mass of the Earth)? This one is straightforward plug and chug from equation 10. First I'll list the quantities: G = 6.67 * 10-11 N m2/kg2 m = mass of jupiter = 1.9 * 1027kg M = mass of sun = 2 * 1030kg r = distance = 5 * 1.5*108km * 1000m/km = 7.5*1011m F = GmM/r2 F = 6.67*10-11N m2/kg2 * 1.9*1027kg * 2*1030kg / (7.5*1011m)2
1 point	F = 4.5*1023N