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Variable Mass Break Up Before the split, momentum is P = MV • M total mass • V center of mass velocity After the split, the sum of momentum is conserved. • P = m1v1 + m2v2 Center of mass velocity remains the same. The kinetic energy is not conserved. v1 M V V v2 Explosions A 325 kg booster rocket and 732 kg satellite coast at 5.22 km/s. Explosive bolts cause a separation in the direction of motion. • K = (1/2)(1057 kg)(5.22 x 103 m/s)2 = 14.4 GJ Find the kinetic energy before separation, and the energy of the explosion. Kinetic energy after separation • K1 = 16.4 GJ • K2 = 0.592 GJ • Satellite moves at 6.69 km/s • Booster moves at 1.91 km/s Kinetic energy before separation is (1/2)MV2 The difference is the energy of the explosion. • Kint = 2.6 GJ Change in Momentum The law of action was redefined to use momentum. dp F dt The change can be due to change in velocity or a change in mass dmv F dt Infinitessimal Change In a short time the following happens: • The mass goes from m to m + Dm • The velocity goes from v to v + Dv • The mass added or removed had a velocity u compared to the object Momentum is conserved (m Dm)v m(v Dv) Dm(v u ) mDv uDm Thrust If there is no external force the force to be applied must be proportional to the time rate of change in mass. m Dv Dm u Dt Dt m dv dm u dt dt The force -u(dm/dt) is the thrust Water Force Thrust can be used to find the force of a stream of water. A hose provides a flow of 4.4 kg/s at a speed of 20. m/s. The momentum loss is dp dm u dt dt • (20. m/s)(4.4 kg/s) = 88 N The momentum loss is the force. Rocket Speed Rockets decrease their mass, so we usually write the mass change as a positive quantity. The equation can be integrated to get the relationship between the mass and increased velocity. dv dm m u dt dt dm dv u m dm vi mi m v f vi u (ln m f ln mi ) vf dv u mf mi v f vi u ln m f Applied Force If there is an external force that must equal the time rate of change in momentum. dv dm Fext m (u v) dt dt Force is needed to maintain the speed. Heavy Water Water is poured into a beaker from a height of 2 m at a rate of 4 g/s, into a beaker with a 100. g mass. There is extra momentum from the falling water. v 2 gh m/t What does the scale read when the water is at 200. ml in the beaker (1 ml is 1 g)? Answer: 302 g dp dm v 2 gh dt dt This is about 0.024 N or an equivalent mass of 2.4 g.