A 1.3-kg booster attached to a 6-kg rocket is initially traveling at a speed of 175 m/s.If the booster is ejected in backward direction with speed of 60 m/s relative to the rocket/booster rest frame,how fast will the rocket then travel?
A)115 m/s
B)162 m/s
C)175 m/s
D)188 m/s
E)235 m/s

Question

Quizplus · Accepted Answer

We can use conservation of momentum to solve this problem.

The momentum of the system (rocket + booster) before the ejection is:

p1 = (m1 + m2)v1

Where m1 is the mass of the rocket, m2 is the mass of the booster, and v1 is the initial velocity of the system.

The momentum of the system after the ejection is:

p2 = m1v2

Where v2 is the velocity of the rocket after the ejection.

The total momentum of the system is conserved, so we can set p1 equal to p2:

(m1 + m2)v1 = m1v2 + m2v2

Simplifying and solving for v2:

v2 = v1(m1 + m2)/(m1 + m2 - m2) - 60

v2 = v1(m1 + m2)/(m1) - 60

v2 = 175(6 + 1.3)/6 - 60

v2 = 188 m/s

Therefore, the rocket will travel at 188 m/s after the booster is ejected. The best choice is D.

A 13-Kg Booster Attached to a 6-Kg Rocket Is Initially Traveling