Chapter 11: Q 82 (page 291)

A two-stage rocket is traveling at 1200 m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 35 m/s relative to the second stage. The first stage is three times as massive as the second stage. What is the speed of the second stage after the separation?

Short Answer

Expert verified

The speed of the second stage after the separation is 1226.25 m/s.

Step by step solution

Step 1. Given information is Velocity of rocket with respect to earth = 1200 m/sRelease of first stage is of 35 m/s with respect to second stage.mass of first stage is 3 times as massive as second stage.

We need to find out the speed of the second stage after the separation .

Step 2. Expressing the total mass of the rocket using principle of conservation of mass

M = m₁ + m₂

m₁is the mass of first stage

m₂ is the mass of second stage

It is given that m₁ = 3m₂

M = 3m₂ + m₂

M = 4m₂

Step 3. Law of conservation of momentum to find the speed of the second stage.

As per law of conservation of momentum,

$\vec{P_{i}} = \vec{P_{f}} M v_{i} = m_{1} v_{1} + m_{2} v_{2}$

Initially rocket is moving with 1200 m/s

And the speed of first stage relative to second stage is given as -35 m/s (negative because they are moving in opposite direction)

So, v₁ - v₂ = -35 m/s

v₁ = -35 m/s + v₂

Putting values in equation,

localid="1649084060349" $M v_{i} = m_{1} (- 35 m / s + v_{2}) + m_{2} v_{2} M v_{i} + (35 m / s) m_{1} = (m_{1} + m_{2}) v_{2} M v_{i} + (35 m / s) m_{1} = M v_{2} v_{2} = v_{i} + \frac{(35 m / s) m_{1}}{M} v_{2} = 1200 m / s + \frac{(35 m / s) (3 m_{2})}{4 m_{2}} v_{2} = 1200 m / s + \frac{105}{4} m / s v_{2} = 1226.25 m / s$