Far out in space, a bored astronaut fires an antique gun at a sack of rocks (mas
ID: 1467649 • Letter: F
Question
Far out in space, a bored astronaut fires an antique gun at a sack of rocks (mass M). The bullet (mass m) bounces around inside the sack a few times, eventually emerging with a reduced, but still high, speed. As the bullet flies off into space, the astronaut remembers with dismay that it was a priceless historical artifact. Fortunately, the astronaut knows the initial speed V0 of the bullet, and that the sack was initially at rest but after being hit was found to be moving with speed vs at angle thetas, as shown in the diagram. In what direction should the astronaut pursue the bullet, assuming that he starts from the initial location of the sack, and what minimum speed vb must he go faster than in order to catch up with it? Use the numerical values given below, and enter the direction as an angle thetab, measured clockwise from the x-axis. m = 0.0250 kg M = 45.0 kg v0 = 1550 m/s v = 0.655 m/s thetas = 29.5degree.Explanation / Answer
oblique
m = 0.025 kg m2 = 45 kg
before collision
speeds
u1x = vo = 1550 m/s u2x = 0
u1y = 0 u2y = 0
after collision
v1x = vs*cos29.5 = 0.655*cos29.5 = 0.57 m/s v2x = vb*costhetab
v1y = -vs*sin29.5 = -0.32254 m/s v2y = v2*sinthetab
from momentum conservation
along y
Piy = Pfy
m1*u1y + m2*u2y = m1*v1y + m2*v2y
0 = -0.025*0.32254 + 45*v2y
v2y = 0.00018 m/s
along x axis
Pix = Pfx
m1*u1x + m2*u2x = m1*v1x + m2*v2x
0.025*1550 = 0.57 + v2x
v2x = 38.18 m/s
v2 = Vb = sqrt(0.00018^2+38.18^2)
vb = 38.18 m/s
directin = tan^(v2y/v2x) = 0
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