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10/25 Momentum in 2D Text: Chapter7 HW: HW Handout “Pendulum and Block” Due Wednesday, 10/30 HW Questions? Examples Impulse Simply the name we give Fnett Another way to find the change in momentum when we know the force and the time Can also find force if we know the change in momentum and the time. Two frictionless rocket sleds are fired from rest toward each other and collide after they both burn out. They stick together after the collision. What is the final velocity of the system? vi = 0 1 m = 60kg Thrust = 50N t = 20s (burn time) 1 1 2 vSYS,f = ? What you know vi = 0 2 m = 30kg Thrust = 25N t = 10s (burn time) 2 What is the total momentum of the system after they collide? The same. No net force so FSYS,net = 0 so pSYS = 0 What is the final velocity of the system after they collide? pSYS = 750kg m/s = mSYSvSYS vSYS = 750/90 = 8.3m/s right, same direction as pSYS Would the answer change if the rockets NO!!!!! Once the collision is done and were still burning and accelerating as the rockets are burnt out, the end result they collided? is the same. Now a third rocket sled is added as shown in the top view below. All three stick together after the collision. What is the final velocity of the system? v =? SYS,f vi = 0 1 m = 60kg NGAS,R = 50N t = 20s (burn time) Fnett = p Our new friend! 1 2 3 vi = 0 3 vi = 0 2 m = 30kg NGAS,R = 25N t = 10s (burn time) m = 45kg NGAS,R = 15N t = 30s (burn time) Can we take advantage of “the system” and find the total impulse on the system? Impulse = the sum of all the individual impulses. = 1000kg m/s right + 250kg m/s left + 450kg m/s up 750kg m/s right Now a third rocket sled is added as shown in the top view below. All three stick together after the collision. What is the final velocity of the system? vSYS,f = ? 450 1 2 3 750 Impulse = the sum of all the individual impulses. = 1000kg m/s right + 250kg m/s left + 450kg m/s up 750kg m/s right pSYS = (750)2 + (450)2 = 875kg m/s vSYS = 875/mTotal = 875/135 = 6.5m/s pSYS,i = 0 in the same direction as pSYS,f pSYS,f = 875kg m/s May seem odd that v is less now but = Arctan (450/750) = 31° above there is more mass in the system! horizontal, to the right Momentum in 2-D A 0.5 kg object moving at constant velocity of 8 m/s east is struck by a hammer. The average force is 300 N directed due north and contact lasts for 0.01 seconds. What is the final velocity (magnitude and direction) of the object? Is the momentum of the object conserved? vi = 8 m/s 0.5 kg No, there is an impulse from an external force. How big is the Impulse? Impulse = Fnett = 3 kg m/s What is the initial momentum? pi = mv = 4 kg m/s What is the final momentum? Momentum in 2-D A 0.5 kg object moving at constant velocity of 8 m/s east is struck by a hammer. The average force is 300 N directed due north and contact lasts for 0.01 seconds. What is the final velocity (magnitude and direction) of the object? y Momentum must be added as vectors. pf vi = 8 m/s x The Impulse is in the y 0.5 kg direction and does not change the momentum in the x direction. pi = mv = 4 kg m/s east pf = 5 kg m/s at 37° above x axis. Impulse = Ft = 3 kg m/s north = py v = 10 m/s A car and a truck, velocities and masses shown, collide and lock bumpers. Find the final velocity (magnitude and direction) of the pair. y vf vc = 15 m/s 370 Mc = 2000 kg vt = 5 m/s Mt = 8000 kg x pt,i,y = ?24,000 kg m/s pc,i,y = 0? psys,f,y = 24,000 ? kg m/s pt,i,x = ?32,000 kg m/s pc,i,x = 30,000 ? kg m/s psys,f,x = ?62,000 kg m/s A car and a truck, velocities and masses shown, collide and lock bumpers. Find the final velocity (magnitude and direction) of the pair. y vf vc = 15 m/s 370 Mc = 2000 kg vt = 5 m/s Mt = 8000 kg x psys,f,y = 24,000 kg m/s psys,f,x = 62,000 kg m/s psys,f = ?66,500 kg m/s vsys,f = ?6.65 m/s = tan ? -1 24,000/62,000 = 21°