. A 50-kg model rocket lifts off by expelling fuel downward at a rateof k = 4.75 kg/s for 10 s. The fuel leaves the end of the rocket with anexhaust velocity of b = −100 m/s. Let m(t) be the mass of the rocketat time t. From the law of conservation of momentum, we find the following differential equation for the rocket’s velocity v(t) (in meters persecond):m(t)v(t) = −9.8m(t) + bdmdt(a) Show that m(t) = 50 − 4.75t kg.(b) Solve for v(t) and compute the rocket’s velocity at rocket burnout(after 10 s)
. A 50-kg model rocket lifts off by expelling fuel downward at a rateof k = 4.75 kg/s for 10 s. The fuel leaves the end of the rocket with anexhaust velocity of b = −100 m/s. Let m(t) be the mass of the rocketat time t. From the law of conservation of momentum, we find the following differential equation for the rocket’s velocity v(t) (in meters persecond):m(t)v(t) = −9.8m(t) + bdmdt(a) Show that m(t) = 50 − 4.75t kg.(b) Solve for v(t) and compute the rocket’s velocity at rocket burnout(after 10 s)
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter9: Linear Momentum And Collisions
Section: Chapter Questions
Problem 34P: A ball of mass 250 g is thrown with an initial velocity of 25 m/s at an angle of 30 with the...
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. A 50-kg model rocket lifts off by expelling fuel downward at a rate
of k = 4.75 kg/s for 10 s. The fuel leaves the end of the rocket with an
exhaust velocity of b = −100 m/s. Let m(t) be the mass of the rocket
at time t. From the law of conservation of momentum, we find the following differential equation for the rocket’s velocity v(t) (in meters per
second):
m(t)v
(t) = −9.8m(t) + b
dm
dt
(a) Show that m(t) = 50 − 4.75t kg.
(b) Solve for v(t) and compute the rocket’s velocity at rocket burnout
(after 10 s)
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