Chapter 8jA missile of mass 100 kg is fired from a plane of ino m/s. The speed o
ID: 2033008 • Letter: C
Question
Chapter 8jA missile of mass 100 kg is fired from a plane of ino m/s. The speed of the missile relative to the plane is 1.00 the plane. Coordinate system: +a is right. 26.L ineat momentum of the plane + missile moving at a g lnitial 1,000sOur goal will he estimaie to estimate the final velocity of A) S.1066 kg-mis B) 1.50E6 kg-mis 47E6 kg-m's .S3E6 kg-mis explain this 1. Whst property of matter helps esx change? A) Energy B) Symmetry C) Scalar D) Inertia 27. Linear momentum of the missile after it is fired A) 1.3E5 kg-m/s 32 Actually, the missile in rcal life would he 3.0E4 kg-m/s C LOES kg-m/s dropped from the plane first before being fired In this case, what would the final velocity of the missile relative to the plane be? 7.0EA kg-ms 28. Let's prepare oursclives to find the final velosity of the plane what is the linear momentum of the plane after the missile is fired? A) 300 m/s B) 1,000 m's c) 0 D) 700 ms 33. Now recompute A) 1.0E2t plane kg-m/s B) 5.IE3 plane ke-m/s A? 700ms B) 300 m/s C) 306 ms D) 286 m/s D) 29. Final velocity of the plane? 34. Is this result more consistent with the idea t momentum is "inertia in action"? 286 m/s ) 280 m/s C) 300 m/s D) 700 m/s A) Yes B) No previous skater + child problem. Are they similar? mass by rn rent from the situation here with that in the missile, we would expect plae to be different from 3 that found in the previous question. how? 30. Since the plane is shedding mass by firing the A) Yes; the plane + missile problem is the skater + child problem in reverse A) Larger B) Smaller C) The same D) Can't say e???.oa?Explanation / Answer
Momentum (P) = Mass * Velocity
m1 = 5,000 kg
u1 = 300 m/s
m2 = 100 kg
u2 = 0 m/s
v2 = 1,300 m/s
P1 = (5,000 kg) * (300 m/s)
P1 = 1,500,000 kg-m/s
P2 = (100 kg) * (0 m/s)
P2f = (100 kg) * (1,300 m/s)
P2f = 130,000 kg-m/s
(26) pi = p1+p2 = 1.50E6 kg-m/s (option B)
(27) p2f = 1.3E5 kg-m/s(option A)
(28) p1f = (m1-m2)v’plane (we subtract m2 since the missile is fired so its weight needs to be deducted)
= (5000-100) v’plane
= 4.9E3 v’plane kg-m/s (option C)
(29)
Conservation of momentum: Pi = Pf
P1 + P2 = P2f + m1v1
(1,500,000 kg-m/s) + (0 kg-m/s) = (130,000 kg-m/s) + [ (4900 kg) * v1]
(1,500,000 kg-m/s) - (130,000 kg-m/s) = (4900 kg) * v1
1,370,000 kg-m/s = (4900 kg) * v1
v1 = (1,370,000 kg-m/s) / (4900 kg)
v1 = 280 m/s (option B)
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