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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