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College Physics (Urone)
Found in: Page 291

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Short Answer

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 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,

\({\rm{Fdt = }}{{\rm{P}}_{\rm{f}}}{\rm{ - }}{{\rm{P}}_{\rm{i}}}\)


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

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 \({\rm{39}}{\rm{.2}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

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