Question

Need help with Q2 ASAP! thanks in advance.

Describe briefly the law of conservation of momentum. In an F vs. r (force vs. position) figure, what does the area under the curve stands for? Problems The "Elysium" is a space station located at an orbit with an altitude 6000 km. It is shaped like a ring and rotates to simulate Earth's gravity. If the radius of the space station is 20 km, what are (a) its rotating (spinning) period (b) its orbiting period? 2. As it is shown in Fig.1. A 1.0-kg object is pressed against the spring (k 30 N/m). 3. A puck is sliding to the right with an initial speed of 20m/s on an air table. The puck 4. If the distance between the Sun and Earth is (-1.50x10" m), the period of compressing it 0.2 m from its equilibrium position. How far up the incline does it travel before coming to rest? ( μ,-0.2) collides with another puck (of the same mass 3 kg), which is initially rest (Fig. 2) Find the final velocities of these two pucks after collision (for elastic collision)? Earth's orbiting around the Sun is T(-365.24 days) Estimate the mass of the Sun in terms of r and T.? (Re 6.37*10 m ME-5.98 10 kg G-667 101" Nm'kg') NASA wants to transport 2000 kg of supply from Cape Canaveral to ISS (International Space Station) whose height is 500 km, how much work is needed? 5. 6. The ballistic pendulum, which consists a large block of wood M-2.0 kg hanging from two cord, can be used to measure the speeds of bullets. A bullet (m 20 g) is fired into the block. The block + bullet then swing upward, and the block of wood rises a vertical distance h 6.0 cm. What is the speed of bullet just prior to the collision? 7. A roller coaster includes a vertical loop of radius 20 m. What is the minimum speed at which the car must move at the top of the loop so that it doesn't lose contact with the track? If the speed of the roller coaster at the bottom of the loop is 20 m/s, what's the apparent weight of the passenger in terms of his weight mg Fig. 1 Fig.2 Befor 53 30° After

Answer 1

m = 1 kg, k = 30 N/m, x = 0.2 m, mu_k = 0.2 Potential energy due to spring U = 1/2 kx^2 This potential energy will convert into kinetic energy So 1/2 kx^2 = 1/2 mv_0^2 v_0^2 = kx^2/m v_0^2 = squareroot kx^2/m = squareroot 30 times 0.04/1 = squareroot 1.2 = 1.09 m/sec a = mg sin theta + f_k/m = mg sin theta + mu_k mg cos theta/m a_net = g sin theta + mu_k g cos theta a = 10 times sin 30 degree + 0.2 times 10 times cos 30 degree a = 5 + 2 times squareroot 3/2 = 5 + 1.732 = 6.73 m/s^2 Now from Newton's 3^rd law of motion v^2 = v_0^2 + 2as a = v_0^2 - 2 times 6.73 times s v_0^2 = 2 times 6.73 times s 1.2 = 2 times 6.73 times s s = 1.2/2 times 6.73 = 0.9 m s = 0.9 m