- #1

- 8

- 0

Hello, could someone please let me know if I have worked this problem out correctly, or if I need to try again? Thank you.

A 40.0 g projectile is launched by the expansion of hot gas in an arrangement showing in Figure P12.4a (attached file). The cross-sectional area of the launch tube is 1.0 cm

A) If the projectile is launched into a vacuum, what is the speed of the projectile as it leaves the launch tube?

B) If instead the projectile is launched into air at a pressure of 1.0 x 10

I chose to use:

Work = the area under a curve

KE = 1/2mV

Area under the curve:

P

P

V

V

ΔP = 1.0 x 10

ΔV = 40 cm

= 3.2 x 10

Area of triangle under curve = (1/2)(-10 x 10

Area of rectangle under curve = (-1.0 x 10

W = -32 J + (-16 J) = -48 J

KE = 1/2mv

v = √(2KE/m)

KE = 48 J

m = 0.040 kg

V = √((2(48J))/0.040 kg) = 48 m/s

B) W = (-1.0 x 10

-32 J/-48 J = 2/3

2/3 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.

## Homework Statement

A 40.0 g projectile is launched by the expansion of hot gas in an arrangement showing in Figure P12.4a (attached file). 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 P12.4b. The values for the initial pressure and volume are P_{i}= 11 x 10^{5}Pa and V_{i}= 8.0 cm^{3}while the final values are P_{f}= 1.0 x 10^{5}Pa and V_{f}= 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?

B) If instead the projectile is launched into air at a pressure of 1.0 x 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?## Homework Equations

I chose to use:

Work = the area under a curve

KE = 1/2mV

^{2}## The Attempt at a Solution

Area under the curve:

P

_{i}= 11 x 10^{5}PaP

_{f}= 1.0 x 10^{5}PaV

_{i}= 8.0 cm^{3}V

_{f}= 40.0 cm^{3}ΔP = 1.0 x 10

^{5}Pa - 11 x 10^{5}Pa = -10 x 10^{5}PaΔV = 40 cm

^{3}- 8.0 cm^{3}= 32 cm^{3}= 3.2 x 10

^{-5}m^{3}Area of triangle under curve = (1/2)(-10 x 10

^{5}Pa)(3.2 x 10^{-5}m^{3}) = -16 JArea of rectangle under curve = (-1.0 x 10

^{5}Pa)(3.2 x 10^{-5}m^{3}) = -32 JW = -32 J + (-16 J) = -48 J

KE = 1/2mv

^{2}v = √(2KE/m)

KE = 48 J

m = 0.040 kg

V = √((2(48J))/0.040 kg) = 48 m/s

B) W = (-1.0 x 10

^{5}Pa)(3.2 x 10^{-5}m^{3}) = -32 J-32 J/-48 J = 2/3

2/3 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.