Question

A 2.58 g particle is moving at 2.51 m/s toward a stationary 6.53 g particle. With what speed does the heavier particle approach the center of mass of the two particles

Answer 1

Given that,

Mass of particle 1, m₁ = 2.58 g = 0.00258 kg

Speed of particle 1, v₁ = 2.51 m/s

Mass of particle 2, m₂ = 6.53 g = 0.00653 kg

Speed of particle 2 = 0 (at rest)

To find,

The speed of centre of mass of the two particles.

Solution,

The formula for the centre of mass of the system is givn by :

`V_(cm)=(m_1v_1+m_2v_2)/(m_1+m_2)\\\\V_(cm)=(0.00258 * 2.51+0.00653 * 0)/(0.00258 +0.00653 )\\\\=0.71\ m/s`

So, the speed of the centre of mass is 0.71 m/s.