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10/25 Momentum in 2D

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 F net t. Another way to find the change in momentum when we know the force and the time

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10/25 Momentum in 2D

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  1. 10/25 Momentum in 2D • Text: Chapter7 • HW: HW Handout “Pendulum and Block” Due Wednesday, 10/30 • HW Questions? • Examples

  2. Impulse • Simply the name we give Fnett • 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.

  3. 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? 1 1 2 2 m = 60kgThrust = 50N t = 20s (burn time) vSYS,f = ? m = 30kgThrust = 25Nt = 10s (burn time) What you know The same. No net force so FSYS,net = 0 so pSYS = 0 pSYS = 750kg m/s = mSYSvSYS vSYS = 750/90 = 8.3m/s right, same direction as pSYS vi = 0 vi = 0 1 2 What is the total momentum of the system after they collide? What is the final velocity of the system after they collide? Would the answer change if the rockets were still burning and accelerating as they collided? NO!!!!! Once the collision is done and the rockets are burnt out, the end result is the same.

  4. 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? 1 1 2 2 vSYS,f = ? 3 Our new friend! 3 Fnett = p 750kg m/s right vi = 0 vi = 0 m = 60kgNGAS,R = 50Nt = 20s (burn time) m = 30kgNGAS,R = 25Nt = 10s (burn time) vi = 0 m = 45kgNGAS,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

  5. 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? 1 2 vSYS,f = ? 3 pSYS = (750)2 + (450)2 = 875kg m/s vSYS = 875/mTotal = 875/135 = 6.5m/s in the same direction as pSYS,f 450  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,i = 0 pSYS,f = 875kg m/s May seem odd that v is less now but there is more mass in the system!  = Arctan (450/750) = 31° above horizontal, to the right

  6. Impulse = Fnett = 3 kg m/s pi = mv = 4 kg m/s 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? No, there is an impulse from an external force. vi = 8 m/s How big is the Impulse? 0.5 kg What is the initial momentum? What is the final momentum?

  7. y x pi = mv = 4 kg m/s east pf = 5 kg m/s at 37° above x axis. pf Impulse = Ft = 3 kg m/s north = py 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? Momentum must be added as vectors. vi = 8 m/s The Impulse is in the y direction and does not change the momentum in the x direction. 0.5 kg v = 10 m/s

  8. 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 x pt,i,y = ? 24,000 kg m/s pc,i,y = ? 0 vt = 5 m/s psys,f,y = ? 24,000 kg m/s pt,i,x = ? 32,000 kg m/s pc,i,x = ? 30,000 kg m/s Mt = 8000 kg psys,f,x = ? 62,000 kg m/s

  9. 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 x psys,f,y = 24,000 kg m/s psys,f,x = 62,000 kg m/s vt = 5 m/s psys,f = ? 66,500 kg m/s vsys,f = ? 6.65 m/s Mt = 8000 kg  = ? tan-1 24,000/62,000 = 21°

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