Physics 101 Fall 2003 Homework Set 4 Exercises: Chapter 6: 2: Padded dashboards make automobiles safer. This is because, if you are in an accident your car will come to an abrupt stop. If you are not wearing your seatbelt, you will come to a stop when you hit the dashboard. Your momentum will change from mv to 0. This means that the dashboard exerted an impulse on you of -mv. Since impulse = Ft, a padded dashboard will lengthen the time of the you-dashboard collision, thereby decreasing the force exerted. 8: The velocity is the same, the mass is the same, therefore the momentum is the same. When gravity changes, only your weight changes, not your mass. 12: No. The gun, having ten times less mass than the bullet, will recoil with a ten times higher speed than the bullet. If the bullet is going fast enough to damage the target that I am aiming at, then the gun will be going fast enough to REALLY hurt when it hits my shoulder. 20: The internal force of the brakes can only stop the automobile if there is an external surface to act upon. Brakes will not slow down your car if it is on a patch of ice or if it is flying through the air. The brakes apply a force to slow down the tires. The tires then apply a force on the road. By Newton's 3rd law, the road applies a reaction force to the tires, slowing down the car. 24: You will not roll backward if go through the motions of throwing the ball but don't actually throw it. You might shift your position backward a little bit, but you will not acquire a backward momentum. This is because of momentum conservation. You start where you and the ball each have zero momentum. If you throw the ball, then the ball has momentum p so you must have momentum -p (same magnitude of momentum but in the opposite direction). If you only pretend to throw the ball, then the ball will still have momentum zero, therefore you must still have momentum zero so your total momentum (yours plus the ball's) still adds to zero. 38: If the bullet squashes and hits the plate, it will have a momentum change of delta p = p(final) - p(initial) = 0 - (mv) = -mv. This is the impulse that the plate exerts on the bullet so it is also equal to the impulse of the bullet on the plate. If the bullet bounces off, it will have a momentum change of delta p = p(final) - p(initial) = m(-v) - (mv) = -2mv. This is because the initial momentum is going in the positive direction but the final momentum is now going in the negative direction. The momentum change is twice as great, so the impulse exerted is twice as great. Bouncing bullets exert a greater impulse on the steel plate. 40: By Newton's third law, the forces on each will be the same. Both will be in the collision for the same period of time. (The truck can't finish colliding before the car.) This means that the impulse = Ft will be the same for both. Since the impulse is the same, both will have the same momentum change. However, since the Escort is lighter, it will have a much larger acceleration (a = F/m). This is what will damage the Escort. 42: When vertically falling sand hits a horizontally moving cart, the cart slows. The first thing to realize is that the vertical motion is irrelvant and does not affect the horizontal motion at all. If it makes thinking about the problem easier, you can consider a single 'block of sand' rather than a steady stream of sand grains. You will have the same results either way. Force picture: Just before the sand lands in the cart, it has zero horizontal velocity. Afterwards, it has a horizontal velocity. Therefore, the cart exerted a horizontal force on the sand to make it go faster. Therefore, by Newton's 3rd law, the sand exerted an equal and opposite force on the cart, making it go slower. Momentum picture: The total horizontal momentum is conserved since there are no outside horizontal forces acting on the cart or the sand. Therefore p(before) = p(after) Now p(before) = m(cart)v(cart) + m(sand)v(sand) = m(cart)v(cart) + 0 p(after) = [m(cart) + m(sand)]*v(both) therefore m(cart)v(cart) = [m(cart) + m(sand)]*v(both) I don't know what the mass of the sand is, but I do know that the mass of the cart with sand in it is greater than the mass of the empty cart. This means that v(both) must be smaller. This problem is identical to a cart on the airtrack hitting and sticking to another stationary cart on the air track. The combined carts move more slowly than the first cart before the collision. Problems: --------- Chapter 6: 2: The car has momentum p = mv = 1000 kg * 20 m/s = 2 * 10^4 kg m/s. You need an impulse that large to make it stop. Therefore Ft = delta p = 2*10^4 kg m/s. If t = 10 s, then you need a force of F = 2*10^4 kg m/s / 10 s = 2 * 10^3 N 4: We are concerned here about the collision between the car and the ground, not with the acceleration due to gravity of the car before it hits the ground. The car hits the ground with momentum p = 1000 kg * 30 m/s = 3 * 10^4 kg m/s a) You want to change the momentum from mv to 0. Therefore the impulse needed to stop it is 3 * 10^4 kg m/s. b) You can not know the force of impact since you do not know the duration of the impact. If the ground is squishy, the duration will be longer and the force less. If the ground is rock hard, the duration will be shorter and the force greater. 8: Again, we use momentum conservation. p(before) = p(after) p(before) = m(engine) v(engine) + m(car)v(car) = 4x * 5 km/h + 0 = 20 x km/h Here I use x to represent the mass of the car and 4x to represent the mass of the engine. You can also plug in a specific value (like 4 kg and 1 kg or 4 tons and 1 ton). p(after) = [m(engine) + m(car)] * v(both) = [4x + x] * v(both) = 5x * v(both) 20 x km/h = 5x * v(both) v(both) = 20 x km/h = 4 km/h --------- 5 x Estimation: ----------- The answer will depend on whether you own a 1 ton Ford Escort or a 2.5 ton SUV. Let's start with the one ton Escort. I have a mass of 80 kg. The Escort has a mass of 1000 kg. Escort: p = 1000 kg 1 mi/h = 1000 kg mi/h (these are horribly mixed units, but it is OK as long as we are consistent) Me: p = 80 kg * v = 1000 kg mi/h v = 1000 kg mi/h / 80 kg = 12.5 mi/h 12 mi/h means one mile in 5 minutes. I can keep up that pace for about a 100 yds. Now we try the SUV. Its mass is 2.5 times larger, so its momentum will be 2.5 times larger, so I will have to run 2.5 times faster. That means that I will need a speed of 2.5 * 12 mph = 30 mph. That's about 15 m/s. Since the world record for the 100 m dash is about 10 s, this is about 50% faster than the fastest human can run. Forget it!