Question

At one instant, the center of mass of a system of two particles is located on the x-axis at x=2.0m and has a velocity of (5.0m/s). One of the particles is at the origin. The other particle has a mass of 0.10kg and is at rest on the x-axis at x= 8.0m. 1.What is the mass of the particle at the origin? 2. Calculate the total momentum of this system. 3. What is the velocity of the particle at the origin?

Answer 1

Let the position of the center of mass be X_cm = 2.0 m

Velocity of the center of mass be V_cm = 5.0 m/s

Let a particle of mass m1 lies at the origin

So it's position is X1 = 0

But it lies at a distance of d1 = 2.0 m from the center of mass

Let the velocity of the mass m1 be V1

Consider a mass m2 = 0.1 kg

It's position relative to origin is X2 = 8.0 cm

But it lies at a distance d2 = 6.0 m from the center of mass

Velocity of the mass m2 be V2 = 0 m/s ( it is at rest )

1) For a center of mass system , we can write

m1 × d1 = m2 × d2

m1 × 2 = 0.1 × 6

m1 = 0.6/2 = 0.3 kg

2) Momentum of the system is simply the momentum of the center of mass.

We know that mass of the system is concentrated at the center of mass. So mass of the system is equal to the mass of the center of mass

Momentum P = mass of the system × velocity of the center of mass

P = ( m1 + m2 ) × V_cm

= ( 0.1 + 0.3 ) × 5.0

= 0.4 × 5.0

= 2.0 kg m/s

3) Momentum of the object is equal to the product of its mass and velocity. Here mass m1 is moving and mass m2 is at rest. So the momentum of the mass m2 is zero . Then the momentum of the system is equal to the momentum of the mass m1.

m1 × V1 = 2

0.3 × V1 = 2

V1 = 2 / 0.3

V1 = 6.66 m/s

Since the mass m1 is less than that of center of mass , it moves with a higher velocity.