Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A 37.0-g projectile is launched by the expansion of hot gas in an arrangement sh

ID: 1525032 • Letter: A

Question

A 37.0-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 P_i = 1.0 times 10^6 Pa and V_i = 8.0 cm^3 while the final values are P_f = 1.0 times 10^5 Pa and V_f = 40.0 cm^3. Friction between the projectile and the launch tube is negligible. If the projectile is launched into a vacuum, what is the speed of the projectile as it leaves the launch tube? 39.45 Your response differs from the correct answer by more than 10%. Double check your calculations, m/s If instead the projectile is launched into air at a pressure of 1.0 times 10^5 Pa, what fraction of the work done by the expanding gas the tube is spent by the projectile pushing air out of the way as it proceeds down the tube? 11.11 Your response differs from the correct answer by more than 10%. Double check your calculations.%

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(10+ 1) (40- 8)10^-6

= 0.5*(11*10^5)*(32*10^-6)

= 17.6 J

according to the work-kinetic energy theoremwe get

W = KE

= (1 / 2) mv^2 - 0

then the speed of the projectile willbe

v = sqrt(2*Wby gas /m)

v = sqrt(2*17.6/(37*10^-3))

v = sqrt(951.35)

v = 30.84 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    

= (1.0*10^5 Pa)*(10^-4 m^2)       

= 10 N

so the energy spent will be

Wspent = Fr s

= Fr(0.32 m)

= 3.2 J

the fraction will be

= Wspent / W

= 3.2/30.84

= 0.104