Question

A comet is traveling through space with a speed of 3.40 ✕ 104 m/s when it collides with an asteroid that was at rest. The comet and the asteroid stick together during the collision process. The mass of the comet is 1.05 ✕ 1014 kg, and the mass of the asteroid is 7.05 ✕ 1020 kg.
a. What is the speed of the center of mass of the asteroid-comet system before the collision?
b. What is the speed of the system's center of mass after the collision?

Answer 1

Answer:

a) vcm = 5.06*10⁻³ m/s

b) vcm = 5.06*10⁻³ m/s

Explanation:

v₁ = 3.40*10⁴ m/s

v₂ = 0 m/s

m₁ = 1.05*10¹⁴ Kg

m₂ = 7.05*10²⁰ Kg

vcm initial = ?

vcm final = ?

a)  Before the collision

vcm = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)

vcm = (1.05*10¹⁴ Kg*3.40*10⁴ m/s + 7.05*10²⁰ Kg*0 m/s) / (1.05*10¹⁴ Kg + 7.05*10²⁰ Kg)

vcm = 5.06*10⁻³ m/s

a)  After the collision

m₁*v₁ + m₂*v₂ = (m₁ + m₂)*vcm

⇒  vcm = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)

vcm = (1.05*10¹⁴ Kg*3.40*10⁴ m/s + 7.05*10²⁰ Kg*0 m/s) / (1.05*10¹⁴ Kg + 7.05*10²⁰ Kg)

vcm = 5.06*10⁻³ m/s

The speed of the system's center of mass is the same value.