A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 98 ms. (a) How far below the release point is the center of mass of the two stones at t = 300 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?
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- Three particles are located on the X-axis. A particle of mass 20 g is located at 0 m. A second particle of mass 20 g is located at 1 m. Where should a particle of mass 26 g be placed so that the Center of Mass of this system is at x=0? (Round your answer to one decimal and enter the units)Enet = 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).
- A tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. To add an extra degree of randomness, the coordinates of the center of mass of the three dice are used as well. The masses of the three dice are 10.10 g, 15.50 , and 18.70 g, and their respective coordinates after one particular throw are (0.2230 m, −0.1850 m), (-0.3270 m, 0.1830 m), and (-0.3730 m, −0.2550 m). What are the resulting coordinates of the center of mass of the dice, xcm and yem? Xcm = m Ycm = mA block of mass 10.0 kg is moving east at 2.0 m/s and another block, of mass 4.0 kg is moving north at 4.0 m/s. The velocity of the center of mass of this two-block system is 2.0 m/s at an angle of 51 degrees south of east 1.8 m/s at an angle of 39 degrees north of east 6.0 m/s at an angle of 51 degrees west of north 1.8 m/s at an angle of 39 degrees east of north 2.0 m/s due westA body of mass 3.1 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/9 of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 3.1 kg body was 5.9 m/s? (a) Number (b) Number Units Units
- A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 54 ms. (a) How far below the release point is the center of mass of the two stones at t = 270 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 Enter your answer for part (a) in accordance to the question statement Units Choose the answer for part (a) from the menu in accordance to the question statement This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times (b) Number Enter your answer for part (b) in accordance to the question statement Units Choose the answer for part (b) from the menu in accordance to the question statement This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or…The vector position of a 3.45 g particle moving in the xy plane varies in time according to i, - (3i + 35)t + 2jr? where t is in seconds and is in centimeters. At the same time, the vector position of a 5.20 g particle varies as i,- 31 - zir? - 6jt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s. x cm cm (b) Determine the linear momentum (in g cm/s) of the system at t = 2.90 s. 9. cm/s (e) Determine the velocity (in cm/s) of the center of mass at t = 2.90 s. cm/s (4) Determine the acceleration (in cm/s) of the center of mass at t = 2.90 s. cm/s? (e) Determine the net force (in µN) exerted on the two-particle system at t = 2.90 s. UN netA system of four particles moves along one dimension. The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system. Find the velocity v4 of particle 4 at t1 = 2.75 s given the details for the motion of particles 1, 2, and 3. particle 1: mj = 1.33 kg, vi (t) = (7.41 m/s) + (0.277 m/s²) × t particle 2: m2 = 3.05 kg, v2 (t) = (8.05 m/s) + (0.567 m/s²) × t particle 3: m3 = 4.25 kg, v3 (t) = (6.33 m/s) + (0.195 m/s²) × t particle 4: m4 = 5.27 kg