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Two particles are moving along the x axis. Particle 1 has a mass m1 and a velocity v1 = +3.2 m/s. Particle 2 has a mass m2 and a velocity v2 = -7.1 m/s. The velocity of the center of mass of these two particles is zero. In other words, the center of mass of the particles remains stationary, even though each particle is moving. Find the ratio m1/m2 of the masses of the particles.

Respuesta :

cjmejiab

Answer:

0.4570

Explanation:

Our values are,

[tex]V_1 = 3.2 m/s\\V_2 = -7.1m/s[/tex]

The center of mass velocity is the sum of each mass's momentum divided by the total mass of the system.

This equation is given by,

[tex]MV = \frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]

We know from the statment that center of mass of these two particles is zero, so

[tex]0 = m_1v_1+m_2v_2[/tex]

Re-arrange to find the ratio,

[tex]m_1v_1 = -m_2v_2[/tex]

[tex]\frac{m_1}{m_2} = -\frac{v_1}{v_2}[/tex]

Substituting,

[tex]\frac{m_1}{m_2} = -\frac{3.2}{-7.1}[/tex]

[tex]\frac{m_1}{m_2} = 0.4570[/tex]

Therefore our ratio is 0.4570.

oladayoademosu

The ratio m₁/m₂ of the masses of the particles is 2.22

Momentum of center of mass

The momentum of the center of mass is given by Mv = m₁v₁ + m₂v₂ where

  • M = combined mass of particles = m₁ + m₂,
  • v = velocity of center of mass = 0 m/s,
  • m₁ = mass of particle 1,
  • v₁ = velocity of particle 1 = + 3.2 m/s,
  • m₂ = mass of particle ₂,
  • v₂ = velocity of particle 2 = -7.1 m/s

Ratio m₁/m₂ of mass of particles

Since v = 0, we have

Mv = m₁v₁ + m₂v₂

M(0) = m₁v₁ + m₂v₂

m₁v₁ + m₂v₂ = 0

m₁v₁ = -m₂v₂

m₁/m₂ = -v₂/v₁

Substituting the values of the variables into the equation, we have

m₁/m₂ = -v₂/v₁

m₁/m₂ = -(-7.1 m/s)/+3.2 m/s

m₁/m₂ = + 7.1 m/s/+3.2 m/s

m₁/m₂ = 2.22

So, the ratio m₁/m₂ of the masses of the particles is 2.22

Learn more about ratio of masses here:

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