This again returns to last week's problem set, and your answers there should help here. For this one, there were two carts on an essentially frictionless track. One cart has a mass of 0.5 kg and the other has a mass of 1.5 kg, and they have velcro that makes them stick together when they collide. They are initially moving toward each other, each with a speed of 0.30 m/s. (a) What is the maximum velocity of the carts just after they collide and stick together? (b) Would this collision be categorized as totally inelastic, inelastic, or (totally) elastic? Explain. (c) How much "convertible kinetic energy" was there in this situation and into what was it converted?

A 1500 kg object moving at 10 m/s collides with a brick wall and comes to a complete stop. If the collision time is 0.05 what was the average stopping force?

show all your steps and write your final answers in a sentence. Please use g = 9.80 for each of these questions to increase precision. Therefore all answers will have 3 significant digits. A ride in the amusement park goes up and down. The ride launches a rider of mass 62.4 kg from an initial velocity of 0.00 m/s[up],to a final velocity of 12.3 m/s[up] and that takes a time of 0.376 seconds. a) Calculate the normal force that the guest experiences from the seat during that time, in which the guest experiences only two forces: normal force and gravity. b) Calculate the g-force the guest experiences during that acceleration. The g-force is the normal force that a person currently experiences, divided by the normal force that the person experiences at rest, with no forces other than normal force and gravity acting on them. Humans have a limit to tolerating g-forces, and high g-forces can be deadly. The most intense rides at a theme park can briefly hit 4-5 g's, while astronauts during…

Consider the diagram which shows a 2D collision between two balls of identical mass (the darker ball is initially at rest). What, if anything, is wrong with this collision? [select ALL of the correct statements from the list below] before 2.0 m/s after 1.89 m/s 0.85 m/s 25° 70⁰ The momentum is not conserved in the horizontal direction The kinetic energy decreases but it should not for this type of collision The momentum is not conserved in the vertical direction The kinetic energy is greater after the collision

From inspection of the torque equation we can find another quantity called the moment of inertia, denoted I, which is the "angular mass". I = mr2 = moment of inertia This was for one point-like mass in the example, but what if we had two balls on the same string some distance apart from one another. Well like before each of these point-like masses has a torque and we are allowed to sum all our torques (similar to forces), i.e. Tnet = T1+ 72 = (mıraı + m2ržaz2)2 Then our moment of inertia for this system would be I = mır} + m2r What if we kept adding masses along the string? Well eventually we would have a distribution of masses along this string that would have to be summed for each radius squared distance they are away from our pivot. If we made a line of masses then we would need to integrate over this distribution of infinitesimal masses to find the total moment of inertia of our system, i.e. Tnet a fr² dm 2 Let's not worry too much about how one would go about integrating out a…

please answer the second one

Please solve 1A and 1B

A 5-kilogram block slides at 20 m/s on a smooth frictionless surface toward a stationary sphere, shown below. The sphere is 4 times the mass of the block. The block strikes the sphere at time t-D0. A plot of the force exerted on the CUBE by the ball as a function of time is shown above right. 20 20 m 10 a) What is the impulse applied to the block? Your answer b) What is the speed of the cube immediately following the collision? (HINT: Impulse = A momentum) Your answer c) What is the velocity of the cube immediately following the collision? (State both direction and magnitude.) Your answe IN

Consider a 50 g mass traveling at 20 m/s impacting and sticking to a 950 g mass that is initially at rest. What is the speed of the combined mass after the collision (in m/s)?

On a very muddy football field, a player of mass m₁ tackles a second one of mass m2 = 3m₁/2. Immediately before the collision, player 1 is slipping with a velocity of v₁ due north and player 2 is sliding with a velocity of v2 = 2v1 due east. 3a. Sketch the situation, placing careful labels, and write formulae for the initial momenta of the two players. Might want to chose a coordinate system that will help you figure out how to add together these momenta. 3b. What is the magnitude of the total momentum of the system of these two players before they collide? 3c. What are the magnitude and direction of the velocity at which the two players move together immediately after the collision? 3d. Calculate the loss of kinetic energy associated with this collision.