A 38.5-g projectile is launched by the expansion of hot gas in an arrangement sh
ID: 1376582 • Letter: A
Question
A 38.5-g projectile is launched by the expansion of hot gas in an arrangement shown in figure a. The cross-sectional area of the launch tube is 1.0 cm2, and the length that the projectile travels down the tube after starting from rest is 32 cm. As the gas expands, the pressure varies as shown in figure b. The values for the initial pressure and volume are Pi = 12 ? 105 Pa and Vi = 8.0 cm3 while the final values are Pf = 1.5 ? 105 Pa and Vf = 40.0 cm3. Friction between the projectile and the launch tube is negligible.
(a) If the projectile is launched into a vacuum, what is the speed of the projectile as it leaves the launch tube?
m/s
(b) If instead the projectile is launched into air at a pressure of 1.5 ? 105 Pa, what fraction of the work done by the expanding gas in the tube is spent by the projectile pushing air out of the way as it proceeds down the tube?
=
Wspent Wenv A 38.5-g projectile is launched by the expansion of hot gas in an arrangement shown in figure a. The cross-sectional area of the launch tube is 1.0 cm^2, and the length that the projectile travels down the tube after starting from rest is 32 cm. As the gas expands, the pressure varies as shown in figure b. The values for the initial pressure and volume are Pi = 12 ? 10^5 Pa and Vi = 8.0 cm^3 while the final values are Pf = 1.5 ? 10^5 Pa and Vf = 40.0 cm^3. Friction between the projectile and the launch tube is negligible. (a) If the projectile is launched into a vacuum, what is the speed of the projectile as it leaves the launch tube? m/s (b) If instead the projectile is launched into air at a pressure of 1.5 ? 10^5 Pa, what fraction of the work done by the expanding gas in the tube is spent by the projectile pushing air out of the way as it proceeds down the tube? Wspent Wenv =Explanation / Answer
(a)
to find the speed first we find the workdoneby the gas on the projectile
the workdone is the area under the curve inthe PV diagram
Wby gas = (triangular area)+(rectangular area)
= (1 / 2) (Po - Pf) (Vf -Vo) + Pf (Vf -Vo)
= (1 / 2) (Po + Pf) (Vf- Vo)
= (1 / 2) 10^5(12+ 1.5) (40- 8)10^-6
= 21.6 J
according to the work-kinetic energy theoremwe get
W = ?KE
= (1 / 2) mv2 - 0
then the speed of the projectile willbe
v = ?(2 Wby gas /m)
= 33.5 m / s
(b)
we know that
and work as
W= (force)(displacement)
pressure = force / area
so the air in front of the projectile wouldexert a retarding force given by
Fr = PairA
= (1.5 x 105Pa) (10^-4 m2)
= 15 N
so the energy spent will be
Wspent = Fr s
= Fr(0.32 m)
= 4.8 J
the fraction will be
Wspent / W = 0.222