Suggested languages for you:

Americas

Europe

Q53PE

Expert-verified
Found in: Page 291

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a $$10,000\;kg$$ ABM that expels $$196\;kg$$ of gas per second at an exhaust velocity of$$2.50 \times {10^3}\;kg$$?

The takeoff acceleration is $${\rm{39}}{\rm{.2}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$.

See the step by step solution

## Step 1: Definition of Acceleration

The rate at which velocity changes is referred to as acceleration.

## Step 1: Given Data

The mass of ABM is $${{\rm{m}}_{\rm{1}}}{\rm{ = 10000}}\;{\rm{kg}}$$.

The expulsion rate is $$\dfrac{{{\rm{dm}}}}{{{\rm{dt}}}}{\rm{ = 196}}\;{\rm{kg/s}}$$.

The exhaust velocity of ABM is $${{\rm{v}}_{\rm{g}}}{\rm{ = 2}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$$.

## Step 2: Calculation of Acceleration

Using the conservation of momentum along vertical direction we get,

$${\rm{Fdt = }}{{\rm{P}}_{\rm{f}}}{\rm{ - }}{{\rm{P}}_{\rm{i}}}$$

So,

\begin{aligned}{\rm{ - mg = m}}\dfrac{{{\rm{dv}}}}{{{\rm{dt}}}}{\rm{ - }}{{\rm{v}}_{\rm{g}}}\dfrac{{{\rm{dm}}}}{{{\rm{dt}}}}\\\dfrac{{{\rm{dv}}}}{{{\rm{dt}}}}{\rm{ = }}\dfrac{{{{\rm{v}}_{\rm{g}}}\dfrac{{{\rm{dm}}}}{{{\rm{dt}}}}{\rm{ - mg}}}}{{\rm{m}}}\end{aligned}

Substituting the values we get,

$$\dfrac{{dv}}{{dt}} = \,a\, = \dfrac{{\left( {{\rm{2}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}} \right){\rm{ \times }}\left( {{\rm{196}}\,{\rm{kg/s}}} \right){\rm{ - }}\left( {{\rm{10000}}\,{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{9}}{\rm{.8}}\,{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {{\rm{10000}}\,{\rm{kg}}} \right)}}$$

$${\rm{ = 39}}{\rm{.2}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$

Hence, the required acceleration is $${\rm{39}}{\rm{.2}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$.