The figure shows a three-particle system, with masses m₁ = 2.5 kg, m2 = 3.9 kg, and m3 = 6.2 kg. What are (a) the x coordinate and (b)the y coordinate of the system's center of mass?y (m)2="0mg1for2mq3x (m)(a)Number1.2936507Unitsm(b)Numberi 1.1Unitsm>

Step 1: Given that m1 = 2.5 kg , m2 = 3.9 kg , m3= 6.2 kg m1 has a coordinates of (0, 0) = (x1,…

Answered: The figure shows a three-particle…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter10: Systems Of Particles And Conservation Of Momentum

Section: Chapter Questions

Problem 21PQ

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A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass ml = 48.0 kg travels in the positive x-direction at 12.0 m/s, and a second piece of mass m2 = 62.0 kg travels in the xy-plane at an angle of 105 at 15.0 m/s. The third piece has mass m3 = 112 kg. (a) Sketch a diagram of the situation, labeling the different masses and their velocities, (b) Write the general expression for conservation of momentum in the x- and y-directions in terms of m1, m2, m3, v1, v2 and v3 and the sines and cosines of the angles, taking to be the unknown angle, (c) Calculate the final x-components of the momenta of m1 and m2. (d) Calculate the final y-components of the momenta of m1 and m2. (e) Substitute the known momentum components into the general equations of momentum for the x- and y-directions, along with the known mass m3. (f) Solve the two momentum equations for v3 cos and v3 sin , respectively, and use the identity cos2 + sin2 = 1 to obtain v3. (g) Divide the equation for v3 sin by that for v3 cos to obtain tan , then obtain the angle by taking the inverse tangent of both sides, (h) In general, would three such pieces necessarily have to move in the same plane? Why?
The vector position of a 3.50-g particle moving in the xy plane varies in time according to r1=(3i+3j)t+2jt2, where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.50 g particle varies as r2=3i2it26jt. At t = 2.50 s, determine (a) the vector position of the center of mass of the system, (b) the linear momentum of the system, (c) the velocity of the center of mass, (d) the acceleration of the center of mass, and (e) the net force exerted on the two-particle system.
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The motion of a human body through space can be modeled as the motion of a particle at the bodys center of mass as we will study in Chapter 9. The components of the displacement of an athletes center of mass from the beginning to the end of a certain jump are described by the equations xj = 0 + (11.2 m/s) (cos 18.5)t 0.360 m = 0.840 m + (11.2 m/s)(sin 18.5)t - 12(9.80 m/s2)t2 where t is in seconds and is the time at which the athlete ends the jump. Identify (a) the athletes position and (b) his vector velocity at the takeoff point. (c) How far did he jump?
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The figure shows a three-particle system, with masses m₁ = 2.5 kg, m2 = 3.9 kg, and m3 = 6.2 kg. What are (a) the x coordinate and (b) the y coordinate of the system's center of mass? y (m) 2 = " 0 mg 1 for 2 mq 3 x (m) (a) Number 1.2936507 Units m (b) Number i 1.1 Units m >