Question

Consider a system of three particles: One particle has mass m1= 2 kg, position r→1= <−5,−1,0> m, and velocity v→1= <3,−2,−5> m/s. The second particle has mass m2= 9 kg, position r→2= <4,2,0> m, and velocity v→2= <8,−10,0> m/s. The third particle has mass m3= 9 kg, position r→3= <4,4,0> m, and velocity v→3= <6,−8,0> m/s. Find the total momentum of the system consisting of all three particles. kg m / s One way to think of systems made of many parts is to separate the motion into two parts. One part is the center of mass motion. This treats the whole system as single object, a point that has the entire mass of the system and is located at the system's center of mass. Find the velocity of the center of mass of the system. m/s The total kinetic energy of the system can be found by adding up all of the individual kinetic energies. This number represents the total kinetic energy the system shares among its different parts. What is the total kinetic energy of this system?

          Consider a system of three particles:
One particle has mass m1= 2 kg, position r→1= <−5,−1,0> m, and velocity v→1= <3,−2,−5> m/s.
The second particle has mass m2= 9 kg, position r→2= <4,2,0> m, and velocity v→2= <8,−10,0> m/s.
The third particle has mass m3= 9 kg, position r→3= <4,4,0> m, and velocity v→3= <6,−8,0> m/s.
Find the total momentum of the system consisting of all three particles.
 kg m / s
One way to think of systems made of many parts is to separate the motion into two parts. One part is the center of mass motion. This treats the whole system as single object, a point that has the entire mass of the system and is located at the system's center of mass.
Find the velocity of the center of mass of the system.
 m/s
The total kinetic energy of the system can be found by adding up all of the individual kinetic energies. This number represents the total kinetic energy the system shares among its different parts.
What is the total kinetic energy of this system?

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Added by Craig O.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

03/30/2023

Step 1

The momentum of each particle is given by the formula: \[ \vec{p} = m \vec{v} \] For particle 1: \[ \vec{p}_1 = m_1 \vec{v}_1 = 2 \, \text{kg} \cdot \langle 3, -2, -5 \rangle \, \text{m/s} = \langle 6, -4, -10 \rangle \, \text{kg m/s} \] For particle Show more…

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Consider a system of three particles: One particle has mass m1= 2 kg, position r→1= <−5,−1,0> m, and velocity v→1= <3,−2,−5> m/s. The second particle has mass m2= 9 kg, position r→2= <4,2,0> m, and velocity v→2= <8,−10,0> m/s. The third particle has mass m3= 9 kg, position r→3= <4,4,0> m, and velocity v→3= <6,−8,0> m/s. Find the total momentum of the system consisting of all three particles. kg m / s One way to think of systems made of many parts is to separate the motion into two parts. One part is the center of mass motion. This treats the whole system as single object, a point that has the entire mass of the system and is located at the system's center of mass. Find the velocity of the center of mass of the system. m/s The total kinetic energy of the system can be found by adding up all of the individual kinetic energies. This number represents the total kinetic energy the system shares among its different parts. What is the total kinetic energy of this system?

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00:01 Hello everyone, we have given that mass m1 is equal to 3 kg, m2 is equal to 6 kg, and m3 is equal to 4kg.

00:10 Now v1 equals to 10 i .kap minus 5j cap plus 14 k kk and v2 vector is equals to minus 14 ikk plus 4 j cap minus 4 k cap and v3 vector is equal to minus 28 i cap plus 38 j cap plus 22 k cap now for the eighth part of the patient the total momentum total momentum of system is p vector which is equals to m1 v1 plus m2v2 vector plus m3 v3 vector.

01:00 On putting the values, we have 3 multiplied by 10 i minus 5 j plus 14 k plus 6 multiplied by minus 14 k plus 6 multiplied by minus 14 k plus 4 j minus 28 i, sorry, huh, 28 i plus 13 ,000, plus 3.

01:29 Now on simplifying we get p vector is equals to minus 166i plus 161 j plus 106k.

01:41 Now for the second part of the question, the velocity of the center of mass is v vector cm is equal to m1 v vector cm is equal to m1 v1.

02:02 Plus m2 v2 plus m3 v3 divided by m1 plus m2 plus m3 therefore the m1 v1 plus m2 v2 with m3 is equal to total momentum hence we get v vector is equal to minus 1665 plus 161 j plus 106 4 divided by 3 plus 6 plus 4 on simplifying we get v vector is equal to minus 12 .57i plus 12 .38 j plus 8 .15k.

02:54 Now for the sixth part of the question, the total kinetic energy, total kinetic energy of this system is kinetic energy, is equal to half m1 v1 square plus half m2 v2 square plus half m2 v2 square plus half n3 v3 square.

03:23 Therefore we had mod of all the velocities.

03:30 Then we have mod of v1 vector is equal to under root 10 square plus minus 5 square plus 14 square which is equal.

03:42 To under root 321.

03:45 Now v2 vector is equal to under root minus14 square plus 4 square plus minus 4 square which is equal to under root of 228.

04:01 Now v3 vector is equal to under root minus 28 square plus 38 square plus 38 square which is equal to under root 2712...

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