The vector position of a 3.85 g particle moving in the xy plane varies in time according to r1 = (3î + 3ĵ)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 r2 = 3î − 2ît2 − 6ĵt. (a) Determine the vector position (in cm) of the center of mass of the system at t = 2.70 s. rcm =       cm (b) Determine the linear momentum (in g · cm/s) of the system at t = 2.70 s. p =       g · cm/s (c) Determine the velocity (in cm/s) of the center of mass at t = 2.70 s. vcm =       cm/s

A set of three masses is arranged as shown below. Calculate the position of the center of mass of the system, i.e., find the coordinates (xcm, Ycm) in units of meters. У (m) 2.00 2.00 kg 5.00 kg 1.00 10.00 kg + x (m) 1.00 2.00

A 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 = 2.99 s given the details for the motion of particles 1, 2, and 3. particle 1: m¡ = 1.21 kg, v1 (t) = (5.87 m/s) + (0.255 m/s?) × t particle 2: m2 = 2.69 kg, v2 (t) = (9.59 m/s) + (0.609 m/s² ) x t particle 3: m3 = 4.37 kg, V3 (t) = (7.87 m/s) + (0.153 m/s²) × t particle 4: m4 = 4.91 kg Ua = m/s

a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length L =40 cm. Find (a) the x coordinate, (b) the y coordinate, and (c) the z coordinate of the center of mass of the box.

In the figure, two particles are launched from the origin of the coordinate system at time t - O. Particle 1 of mass m₁ - 5.20 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.7 m/s. Particle 2 of mass m₂ -4.50 g is shot with a velocity of magnitude 15.6 m/s, at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height Hmax reached by the com of the two-particle system? In unit-vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches Hmax? (a) Number 3.05 (b) Number 10.7 (c) Number 0 Units i+ 0 m 3.3 j Units m/s j Units m/s^2

A system composed of the following particles: a 0.500 kg particle at the origin; a 1.25 kgkg particle at x = 0.150 mm, y = 0.200 mm; and a 0.750 kg particle at x = 0.200 mm, y = -0.800 mm. a.  Find the x-coordinate of the center of mass of the system. b. Find the y-coordinate of the center of mass of the system.

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. µN

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Three ice skaters meet at the center of a rink and each stands at rest facing the center, within arm's reach of the other two. On a signal, each skater pushes himself away from the other two across the frictionless ice. After the push, skater A with mass mA = 85.0 kg moves in the negative y-direction at 3.00 m/s and skater B with mass mB = 95.0 kg moves in the negative x-direction at 4.00 m/s. Find the x- and y-components of the 90.0 kg skater C's velocity (in m/s) after the push. vCx =   vCy =

A 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