Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.
The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;
K.E₁ + P.E₁ = K.E₂ + P.E₂
[tex]\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B[/tex]
[tex]v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left[/tex]
At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.
The complete question is below:
Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?
Learn more about conservation of energy here: https://brainly.com/question/166559
Answer:
Explanation:
the linear momentum is conserved MiVi=MaVa+MbVb
MV0=2V0M/3+MVb
Vb=V0/3
