Consider a system consisting of three particles m1 3 kg 1 7...

Consider a system consisting of three particles: m1 = 3 kg, 1 = < 7, -6, 15 > m/s m2 = 3 kg, 2 = < -15, 6, -6 > m/s m3 = 3 kg, 3 = < -21, 35, 20 > m/s (a) What is the total momentum of this system? tot = kg · m/s (b) What is the velocity of the center of mass of this system? cm = m/s (c) What is the total kinetic energy of this system? Ktot = J (d) What is the translational kinetic energy of this system? Ktrans = J (e) What is the kinetic energy of this system relative to the center of mass? Krel = J

Transcript

00:01 Fellow students, we are given width, we are given with mass is equals to 3kg and velocity vector is equals to 7 i cap minus 6 j cap plus 15 k cap meter per second and which is equal to its magnitude is equals to under root 310 meter per second.

00:37 Now we are given with v2 actor is equal to minus 15 j cap minus 15 i cap sorry minus 15 i cap plus 6 j gap minus 6 k gap meter per second in mass 2 is also equals to 3 kgs and our velocity magnitude here is v2 is equal to root 200 7 meter per second meter per second and we are given with mass 3 is also equals to 3 kgues 3 k velocity 3 vector is equals to minus 21 i cap plus 35 j cap 35 j cap minus 20 k cap 20 k cap meter per second and magnitude of v3 will be equals to 2006 meter per second now we need to first for first we need to find out the total momentum of the system total momentum of the system which is equals to momentum sum or vector sum vector sum of individual momentum individual momentum of each mass member individual momentum of each mass which is b vector total is m1 b1 vector plus m2 v2 vector plus m 3 v3 vector which is equal to 3 multiplied by 7 i cap minus 6 j cap plus 15 k cap plus 3 multiplied by minus 15 i cap plus 6 j cap minus 6 k cap is plus 3 multiplied by minus 21 i cap plus 35 j cap minus 25 j cap minus 20 k cap which is equals to 87 i cap minus 87 i cap plus 105 j cap minus 33 kg meter per this is our solution next we need to find out velocity of center of mass velocity of center of mass which is equal to v center of mass is equal to m1 v1 vector plus m2 v2 vector plus m3 v2 divide by m1 plus m2 plus m3 therefore putting the values we get v center of mass equals to b vector divide by m1 plus m2 which is equals to minus 87 i kip plus 105 j cap minus 33 k kk divide by 3 plus 3 plus 3 which is equals to minus 29 3 i cap plus plus 35 divided by 3 j k k k k kp meter per second so this is our answer now we need to find total of the system so total kinetic energy of the system will be total kinetic energy of the system will be total kinetic energy of the system that is kinect.

05:59 Total is equal to half m1 wave in square plus half m2 v2 square plus half m3 v3 squared putting the values we get half 310 we'll take out the three common that is mass which is common in all of them half multiplied by 3 multiplied by 310 plus 297 plus 266 which is equals to 3 by 2 divided by 2 multiplied by 2 ,673 joules which is equals to 4 ,0409 .5 joules.

07:12 Now we need to find translational kinetic energy or system.

07:19 Translational kinetic energy of the system.

07:34 So this is the same as the k .e total because our motion is translational only.

07:41 Therefore, k .e.

07:45 Translational is equals to 4 ,09 .5 jule only.

07:53 Now we need to find relative velocity of each particle.

08:02 Relative velocity of each particle.

08:09 And that can be found out by the velocity of particle minus by velocity of center of mass...