Page 105 - Schaum's Outline of Theory and Problems of Applied Physics

90 MOMENTUM [CHAP. 8

The airplane’s initial velocity is v 1 = 0, and its final velocity is
1000 m/km

v 2 = (700 km/h) = 19.4 m/s
3600 s/h
Hence Impulse = momentum change
Ft = m(v 2 − v 1 ) = mv 2
mv 2 (50,000 kg)(19.4 m/s)
t = = = 139 s = 2 min 19 s
F 70,000 N
SOLVED PROBLEM 8.5

A 2400-lb car strikes a fence at 30 ft/s (about 20 mi/h) and comes to a stop in 1 s. What average force
acted on the car?
The initial and final velocities of the car are 30 ft/s and 0, respectively. Hence

Impulse = momentum change
w
Ft = m(v 2 − v 1 ) = (v 2 − v 1 )
g
w(v 2 − v 1 ) (3200 lb)(0 − 30 ft/s)
F = = =−3000 lb
2
gt (32 ft/s )(1s)
The minus sign means that the force which acted to stop the car is in the opposite direction to its initial velocity.

SOLVED PROBLEM 8.6
The thrust of a rocket is the force developed by the expulsion of its exhaust gases. (a) Find the thrust of a
rocket that uses 30 kg/s of fuel and whose exhaust gases leave the rocket at 3 km/s. (b) If the initial mass
of the rocket is 5000 kg of which 3500 kg is fuel, find its initial and final accelerations.
(a) When a mass m of exhaust gas leaves a rocket in the time interval t, its momentum change (mv) is

(mv) = v m

Since the impulse Ft given to the rocket equals the momentum change of the exhaust gases,
F t = (mv)
(mv) m
F = = v
t t
The thrust of a rocket is the product of the exhaust speed and rate at which fuel is consumed. Here v = 3 km/s =
3000 m/s and m/ t = 30 kg/s, so
m
4
F = v = (3000 m/s)(30 kg/s) = 9 × 10 N = 90 kN
t
(b) The initial mass of the rocket is m 0 = 5000 kg, and the initial force available for its acceleration is F − m 0 g.
Hence
2
4
F − m 0 g 9 × 10 N − (5000 kg)(9.8 m/s ) 2
a 0 = = = 8.2 m/s
m 0 5000 kg
The final mass of the rocket is 1500 kg. Hence its final acceleration before it runs out of fuel is
2
4
F − mg 9 × 10 N − (1500 kg)(9.8 m/s )
a = = = 50.2 m/s 2
m 1500 kg