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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

01

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 .

02

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

03

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

As per law of conservation of momentum,

Pi鈫�=Pf鈫�Mvi=m1v1+m2v2

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" Mvi=m1(-35m/s+v2)+m2v2Mvi+(35m/s)m1=(m1+m2)v2Mvi+(35m/s)m1=Mv2v2=vi+(35m/s)m1Mv2=1200m/s+(35m/s)(3m2)4m2v2=1200m/s+1054m/sv2=1226.25m/s

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

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Most popular questions from this chapter

Section 11.6 found an equation for vmax of a rocket fired in deep space. What is vmax for a rocket fired vertically from the surface of an airless planet with free-fall acceleration g? Referring to Section 11.6, you can write an equation for 鈭哖y, the change of momentum in the vertical direction, in terms of dm and dvy. 鈭哖y is no longer zero because now gravity delivers an impulse. Rewrite the momentum equation by including the impulse due to gravity during the time dt during which the mass changes by dm. Pay attention to signs! Your equation will have three differentials, but two are related through the fuel burn rate R. Use this relationship鈥攁gain pay attention to signs; m is decreasing鈥攖o write your equation in terms of dm and dvy. Then integrate to find a modified expression for vmax at the instant all the fuel has been burned.

a. What is vmax for a vertical launch from an airless planet ? Your answer will be in terms of mR, the empty rocket mass; mF0, the initial fuel mass; vex, the exhaust speed; R, the fuel burn rate; and g.

b. A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of thrust. What is this rocket鈥檚 speed at the instant all the fuel has been burned if it is launched in deep space ? If it is launched vertically from the earth?

A 50kgarcher, standing on frictionless ice, shoots a 100g arrow at a speed of 100m/s. What is the recoil speed of the archer?

A spaceship of mass 2.0 脳 106 kg is cruising at a speed of 5.0 脳 106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0 脳 105 kg, is blown straight backward with a speed of 2.0 脳 106 m/s. A second piece, with mass 8.0 脳 105 kg, continues forward at 1.0 脳106 m/s. What are the direction and speed of the third piece?

At what speed do a bicycle and its rider, with a combined mass of 100kg, have the same momentum as a 1500kg car traveling at 5.0m/s?

(3000kg)vfx=(2000kg)(5.0m/s)+(1000kg)(-4.0m/s)

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem, including a pictorial representation.
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