Physics 101 Lawrence Weinstein Homework set 4 Solutions All problems are from Chapter 6 1-Steps: ------------------- 6-2: p = mv = 50 kg * 4 m/s = 200 kg-m/s 6-8: Momentum is conserved. Therefore the initial momentum equals the final momentum. p_i = 2 kg * 3 m/s + 4 kg * 0 m/s = 6 kg-m/s p_f = (m1 + m2) * v = p_i v = p_i / (m1 + m2) = 6 kg-m/s / (2 kg + 4 kg) = 1 m/s Exercises: ----------------------- 6-4: When the gymnast hits the mat with a certain speed, she will have a certain change in momentum. To make her momentum change, the mat exerts a certain impulse on her. Impulse = Ft. In order to minimize the force (to reduce the damage she suffers in the collision with the mat), she wants to make the time or duration of the collision as large as possible. The best way to do that is with a nice soft thick mat. 6-8: A person can only safely absorb a certain amount of force, F_maximum. The softer the surface, the longer the duration of the collision with it, therefore the larger the total impulse that can safely be exerted by the surface I_max = F_max * t, therefore the larger the momentum the person can safely hit the surface with. 6-12: p = mv depends only on mass, not on weight. Your mass does not change. Only the gravitational force on you changes when you are on the Moon. 6-24: The external force is exerted between the tires and the road. This slows down the car. If there is no friction between the tires and the road (eg: if you are on ice), then there will be no external force and you will not slow down. 6-28: There are (at least) two explanations for this: a) momentum is conserved. The initial momentum is p_before = p_you + p_ball = 0. Since p_after = p_before, afterwards you must have that p_you = - p_ball. If you do not throw the ball, then the only way for p_you + p_ball = 0 is for both of you to not be moving. b) When you pretend to throw the ball, you can consider this a combination of two motions. First you exert a force on the ball to increase its speed, then you exert a force on the ball to decrease its speed back to zero. In the first half, the ball exerts a backward force on you and in the second half, the ball exerts a forward force on you. The total impulse you exert on the ball is zero, therefore the total impulse that the ball exerts on you is zero. 6-44: The bouncing bullet exerts a greater force on the plate. This is because the bouncing bullet has a greater change in momentum than the bullet that sticks (because momentum is a vector -- we did this as a quiz in class). The bouncing bullet has a greater momentum change, therefore the plate exerted a greater impulse on the bouncing bullet to make its momentum change, therefore the bouncing bullet exerted a greater impulse on the plate (by Newton's 3rd law). 6-48: The forces on the two vehicles will be equal and opposite. (Newton's 3rd Law) Since the two forces are exerted for the same time (ie: the time that the two vehicles are in contact with each other), the Impulse = Ft will be the same. Since the impulse on the two vehicles is the same, then their change in momentum will be the same. (since Ft = delta p) The Escort will have a much larger change in velocity, since delta v = delta p / m and it has a much smaller mass than the truck. This means that it will have a much larger acceleration since a = delta v / t. You must give a reason for each of your answers. Problems: ------------------------------- 6-2: a) use impulse = Ft = delta p so F = delta p / t delta p = p_after - p_before = 0 - (8 kg * 2 m/s) = - 16 kg-m/s therefore F = -16 kg-m/s / 0.5 s = -32 N Note that the minus sign indicates that the force is opposite in direction to the momentum. You do not need the minus sign in your answer to be correct. b) 32 N 6-4: The momentum change of the car during its collision with the ground is 1000 kg * 30 m/s = 3*10^4 kg-m/s. a) the impulse on the car during the collision with the ground can be calculated since it equals the momentum change = 3*10^4 kg-m/s (you do not need to give the numeric value of the answer) b) the force is not knowable since we do not know the duration of the collision 6-6: momentum is conserved p_after = p_before p_before = (40 kg * 0 m/s) + (15 kg * 3 m/s) = 45 kg-m/s p_after = (m1 + m2) * v so that v = p_after / (m1 + m2) = p_before / (m1 + m2) v = 45 kg-m/s / (15 kg + 40 kg) = 45 kg-m/s / (55 kg) = 0.82 m/s (Note that 2 or 3 significant figures is OK. If you write 4 or more (eg: 0.818181), you should lose a point.) Estimation: --------------------------------- Estimate the momentum of a drifting continent. (Hint: you can estimate the speed of the continent from knowing that the Atlantic Ocean opened up to its present width over the last 100 million years.) Express your answer in standard units (kg-m/s). This problem was more complicated than the previous ones. We need the velocity and mass of a drifting continent. Let's use North America. v = d / t It moved the width of the Atlantic Ocean in 10^8 years. This means that d = 3000 miles = 5000 km and t = 10^8 yr * (3 * 10^7 s / 1 yr) = 3*10^15 s. Thus v = d / t = 5*10^6 m / 3*10^15 s = 2*10^(-9) m/s. Not very fast. Now we need the mass. Let's start with the volume. V = length * width * height (or depth). The distance from here to California is 3000 mi = 5000 km. The distance from northern Canada to Mexico is about the same. The continent is a geological plate. The depth of this is much more than 1 mile (since mines go down further than that) but less than 1000 miles, so we'll take the average as 30 miles. (If it's between 1 and 1000, then we can say it is between 10 and 100. I choose 30 since 30 = 10*3 and 30 = 100/3, so it is geometrically in the middle.) Thus Volume V = 5000 km * 5000 km * 50 km = 10^9 km^3 * (1000 m / 1 km)^3 V = 10^18 m^3 Now we need to calculate the mass from the volume. 1 pint of water weighs about a 1 lb and 1 liter of water (or soda) has a mass of 1 kg. In the back of the book, it shows that 1 liter = 10^(-3) m^3. Thus 1 m^3 = 10^3 l so that 1 m^3 of water has a mass of 1000 kg. Rock is denser than water so that 1 m^3 of rock has a mass of about 3*10^3 kg. Now we can put it all together. m = density * V = (3*10^3 kg/m^3) * 10^18 m^3 = 3*10^21 kg Rather a lot. p = mv = 3*10^21 kg * 2*10^(-9) m/s = 6*10^12 kg-m/s Note that that is about the same as 1000 supertankers or 1000 aircraft carriers moving at 10 m/s or about 20 mph.