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 t¡ = 2.83 s given the details for the motion of particles 1, 2, and 3. particle 1: m¡ = 1.21 kg, vi (1) = (5.65 m/s) + (0.299 m/s² )×t particle 2: m2 = 2.93 kg, v2 (t) = (8.93 m/s) + (0.357 m/s²) × t particle 3: m3 = 4.73 kg, v3 (t) = (7.65 m/s) + (0.321 m/s² ) ×t particle 4: m4 = 4.55 kg
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- 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 ?4v4 of particle 4 at ?1=3.07 st1=3.07 s given the details for the motion of particles 1, 2, and 3. particle 1:?1=1.33 kg,?1(?)=(6.97 m/s)+(0.321 m/s2)×?particle 1:m1=1.33 kg,v1(t)=(6.97 m/s)+(0.321 m/s2)×t particle 2:?2=3.53 kg,?2(?)=(8.27 m/s)+(0.315 m/s2)×?particle 2:m2=3.53 kg,v2(t)=(8.27 m/s)+(0.315 m/s2)×t particle 3:?3=3.89 kg,?3(?)=(7.43 m/s)+(0.447 m/s2)×?particle 3:m3=3.89 kg,v3(t)=(7.43 m/s)+(0.447 m/s2)×t particle 4:?4=4.55 kgthis is a multi step problem but I am just having trouble with this part of the problem. Here are the values that I correctly calculated: mass of particle 1=6 kg initial velocity for particle 1= 185i+138j final velocity for particle 1= -68.89i+-50.67j F21=-15.9e-t/96i+-100sin(2πt/96)j I am trying to find: What is the x displacement for particle 1? I think that I need to use Newtons second law to get an expression for acceleration, and then integrate but I am not getting the right answers.A 0.25 kg mass of clay is moving with a horizontal velocity of v when it collides and sticks to a 5 kg object hanging from the end of a 1 meter long rod. Find the minimum velocity v so that the objects will make one complete revolution.
- A marble with mass mA = 2.5 [g] moves eastward with speed VA = 1.75 [m/s]. It then collides with another marble of mass mg = 1.25 [g] and is moving northward with a speed VB = 2.30 [m/s]. After the collision, marble A now moves at 50° north of east, while marble B moves at 20° south of east. 1 A. What is the speed of marble A after the collision? B. What is the speed of marble B after the collision? C. Is energy lost or gained in the system?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 kgConsider the collision pictured below. While m2 = 10kg is initially at rest, horizontally moving mị = 6kg with a speed of v1 = 11.8m/s collides with it. After the collision, the two masses scatter with speeds of v, = 7.8m/s and v,. If 01 = 48°, determine the speed of m2 after the collision, v, =?. Express your answer in units of m/s with one decimal place. Note: Please note that we do not know whether or not this is an elastic collision. Just before the collisbn Just after the collision at rest Answer:
- 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/sYou are analyzing the possible defenses for an asteroid that is going to crash into Earth.Initially, the asteroid of mass ma is traveling to the left with speed vai . A missile of mass mm with initial velocity vmi to the right collides with the asteroid, and embeds itself inside the asteroid. Find the common final velocity of the missile and asteroid after the collision. ma = 1.3000E+9 kgmm = 9.1000E+7 kgvai x = -2200 m/svmi x = 12571 m/s [the final velocity for this one was -1233.67m/s] = vf x The missile then explodes, causing the asteroid to break into two chunks of mass m1 and m2. Solve algebraically for the final speeds of pieces 1 and 2 in terms of the masses of the two pieces (m1 and m2), the angles, the total mass before the explosion, and vf x (the velocity after the collision that you found above).Then use the following values for the parameters to get values for the speeds: m1 = 2.7820E+8 kgm2 = 1.1128E+9 kgθ1 = 30.2 degreesθ2 = 52.5 degrees This problem is dealing with…Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as m₁ = 0.0145 kg, m2 = 0.0141 kg, m3 = 0.0245 kg, and m4 = 0.0105 kg. At this very instant, the x and y components (Ux, U,) of the mice's velocities are, respectively, (U₁x, U₁₁y) = (0.591 m/s, -0.425 m/s,) (v2.x, U2,y) = (-0.605 m/s, -0.933 m/s,) (v3,x, U3,y) = (0.259 m/s, 0.305 m/s), and (V4x, V4,y) = (-0.211 m/s, 0.569 m/s). Calculate the x and y components px and py of Bryce's mice's total momentum. Px = kg.m/s Py = kg.m/s
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