The 10-kg mass A is moving at 5 m/s when it is 1m from the stationary 10-kg mass B. The coefficient of kinetic friction between the floor and the two masses is μ = 0.6, and the coefficient of restitution of the impact is e=0.5. Determine how far B moves from its initial position as a result of the impact.
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- A metal object of mass m = 45.77 g with sharp edges hits a wooden block of mass M = 425.3 g at rest on a horizontal surface. Assume that the coefficient of kinetic friction between the block and the surface μk is 0.43. When the metal object hits the block, it is stuck with the block and both start to slide together with velocity vf. They move together after the collision, and stop after sliding ∆x = 1.95 m. What is the initial velocity vi of the metal object before it hits the block in m/s? Keep three significant figures for the answer.A stone is dropped at t = 0. A second stone, with 4 times the mass of the first, is dropped from the same point at t = 55 ms. (a) How far below the release point is the center of mass of the two stones at t = 500 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time? (a) Number Units (b) Number i Units >A rocket, which is in deep space and initially at rest relative to an inertial reference frame, has a mass of 86.2 × 105 kg, of which 14.7 × 105 kg is fuel. The rocket engine is then fired for 180 s, during which fuel is consumed at the rate of 480 kg/s. The speed of the exhaust products relative to the rocket is 2.98 km/s. (a) What is the rocket's thrust? After the 180 s firing, what are (b) the mass and (c) the speed of the rocket?
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- Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite................m/sEnet = The vector position of a 3.80 g particle moving in the xy plane varies in time according tor, = (3î + 3j)t + 2ĵt2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.35 g particle varies as 2 sr, = 3î – 2ît? - 6jt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.40 s. rcm 4.744i – 7.38j cm (b) Determine the linear momentum (in g• cm/s) of the system at t = 2.40 s. p = 11.4i – 35.58j g• cm/s (c) Determine the velocity (in cm/s) of the center of mass at t = 2.40 s. 1.245i – 3.888j cm/s = cm (d) Determine the acceleration (in cm/s2) of the center of mass at t = 2.40 s. 0.6775j cm/s? = cm (e) Determine the net force (in µN) exerted on the two-particle system at t = 2.40 s. µNProblem 3: A particle with mass ma = 3.00 kg is located at ra = (2.50 i + 3.50 j) m, and a second particle of mass m2B = 5.00 kg is located at rB = (1.50 i - 3.00 j) m. Find the location of the center of mass of the system relative to the point (1,1).
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