Question

A two-stage rocket moves in space at a constant velocity of +4390 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1340-kg upper stage is +5600 m/s. What is the velocity (magnitude and direction) of the 2620-kg lower stage immediately after the explosion

Answer 1

Mass of upper stage of the rocket = m₁ = 1340 kg

Mass of the lower stage of the rocket = m₂ = 2620 kg

Total mass of the two-stage rocket = M

M = m₁ + m₂

M = 1340 + 2620

M = 3960 kg

Initial velocity of the two-stage rocket = V = 4390 m/s

Velocity of the upper stage after the explosion = V₁ = 5600 m/s

Velocity of the lower stage after the explosion = V₂

By conservation of linear momentum,

MV = m₁V₁ + m₂V₂

(3960)(4390) = (1340)(5600) + (2620)V₂

V₂ = 3771 m/s

Velocity of the lower stage immediately after the explosion = +3771 m/s