Chapter 9Page 1CHAPTER 9 - Linear Momentum and Collisions1.We find the force on the expelled gases fromF= ∆p/∆t= (∆m/∆t)v= (1200 kg/s)(50,000 m/s) = 6.0×107N.An equal, but opposite, force will be exerted on the rocket: 6.0×107N, up.2.For the momentum p= 4.8t2i– 8.0j– 8.9tk, we find the force fromF= dp/dt = 9.6ti– 8.9k, in SI units.3.(a)p= mv= (0.030 kg)(12 m/s) = 0.36 kg·m/s.(b)The force, opposite the direction of the velocity, changes the momentum:F= ∆p/∆t;– 2.0×10–2N = (p2– 0.36 kg·m/s)/(12 s), which gives p2= 0.12 kg·m/s.4.The change in momentum is∆p = p2– p1= mvj– mvi= (0.145 kg)(30 m/s)j– (0.145 kg)(30 m/s)i= – (4.35 kg·m/s)i+ (4.35 kg·m/s)j.5.The force changes the momentum:F= (26 N)i– (12 N/s2)t2j= dp/dt.Because Fis a variable force, we integrate to find the momentum change:dp=Fdt;∆p=(26N)i– (12N/s2)t2jdt1.0s2.0s= (26N)ti– (4.0N/s2)t3j1.0s2.0s=(26N?s)i– (28 N?s)j.6.If Mis the initial mass of the rocket and m2is the mass of theexpelled gases, the final mass of the rocket is m1= M– m2.Because the gas is expelled perpendicular to the rocket in therocket’s frame, it will still have the initial forward velocity,so the velocity of the rocket in the original direction will notchange. We find the y-component of the rocket’s velocity afterfiring fromv1⊥= v0tan = (120 m/s) tan 23.0° = 50.9 m/s.Using the coordinate system shown, for momentumconservation in the y-direction we have0 + 0 = m1v1⊥– m2v2⊥, or(M– m2)v1⊥= m2v2⊥;(4200 kg – m2)(50.9 m/s) = m2(2200 m/s), which gives m2= 95 kg.7.(a)We choose downward as positive. For the fall we havey= y0+ v0t1+ !at12;h= 0 + 0 + !gt12, which gives t1= (2h/g)1/2.To reach the same height on the rebound, the upward motion must be a reversal of the downwardv0Beforev2v1Afterxygas
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