Problems ? When Old Mixes With New P2-4. A 32 g bullet is fired from a rifle at

Question

Problems ? When Old Mixes With New P2-4. A 32 g bullet is fired from a rifle at 380 m/s into a water bottle at pointblank range. The bullet travels completely through the 12 cm long bottle, located 2 m above the ground. It finally comes to earth at a point 3 m downrange from the point where it left the water. Find A) the average force exerted by the water on the bullet and B) the total time elapsed from when the rifle is fired to when the bullet hits the ground. P2-5. A 1.5 x 105 kg rocket is burning fuel and, as a result, losing mass at a rate of 30 kg/s. The rocket also constantly accelerates upward from rest to 27 m/s in 9 minutes. Assuming the rocket started from rest at the ground, determine the net force acting on the rocket after 2 minutes.

Explanation / Answer

2.4) We need the horizontal velocity as the bullet leaves the bottle of water.

Horizontal velocity is independent of vertical velocity ===> To get the time of flight from the bottle we figure out how long to fall 2 m

d = 0.5at^2
2 m = 0.5 * 9.8 m/s^2 * t^2
t^2 = 2/4.9 = 0.41 s
t = 0.64 s

===> We now know the bullet went 3 m in 0.64 s
d = Vh*t
3 m = Vh * 0.64
Vh = 4.69 m/s

Bullet lost 380 - 4.69 = 375.31 m/s over 12 cm = 375.31 m/.s over 0.12 m

Now we balance Energy/Work

KE before = 0.5MV^2 = 0.5 * 0.032kg * (380 m/s)^2 = 0.5 * 0.032 kg * 144,400 = 2310.4 J

KE after = 0.5 * 0.032 * (4.69 m/s)^2 = 0.352 J

Energy Lost: 2310.4 - 0.35 = 2309.95 J

Work = Force * distance = KE lost

0.12 m * F = 2309.95
F = - 19,249.57 N (That's A)

Assuming 0 time rifle to bottle

Vf = at * Vo
4.69 = -19249.57 *t + 380
-19249.57 t = - 380 + 4.69
t = 0.02s

Total time: 0.02s + 0.64 = 0.66s

2.5) dF = dm a
F = a ?dm
F = a (mf - mi)
a = 27/(9*60) = 5x10^-2 m/s^2
mi = 1.5x10^5
mf - mi =30*?t = 30*120 = 3.6x10^3 kg
then
F = 5x10^-2*3.6x10^3 = 180 N

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