Question

Can someone help with 2, 3 and 4 please

1. 2. An arrow with a mass of 115 g and a velocity of 50.0 m/s hits and sticks to a piece of wood of mass 2.10 kg. If the block of wood is on a frictionless, horizontal surface, with what velocity will the block move? A 2.50 kg mass moving 7.50 m/s to the right collides head on with a 4.90 kg mass. After the collision the 2.50 kg mass is moving 5.00 m/s to the left and the 4.90 kg mass is moving 4.88 m/s to the right. a. Calculate the velocity of the 4.90 kg mass before the collision b. If the collision lasts for 0.0625 seconds calculate the force that acted on each mass during the collision. An 8.00 kg shell initially moving at 28.8 m/s (East) explodes into 3 fragments. A 3.00 kg fragment flies directly north at 50.0 m/s and a 2.00 kg fragment flies directly east at 40.0 m/s. Calculate the velocity of the third fragment A 475 kg speedboat is racing 31.0 m/s [N 35 E) and crashes into a stationary 180 kg catamaran. After the collision the speedboat has a velocity of 18.0 m/s (N 67° E). Determine the final velocity of the catamaran. 3. 4. C. APPLICATION 1. Select a technology that applies the physics principles of momentum and impulse. (Ex: air bags, seatbelts, head rests, crumple zones in cars, design of median or side barriers on roadways, helmets and concussions, bungee jumping, golf, rocket propulsion, the karate chop, etc.) Use your knowledge of momentum and impulse to explain how the technology works in 1 to 2 paragraphs.

Answer 1

Betone collision 2 50kg +7.50m/s After wrision 2-sokg 490 kg) v (4.90kg 4.88 m/s. 5 m/s (a) Applying momentum conourvation prinuple 2-50x7.5+ 4.9 x V = 4.90x4.88 - 5x2.5 mass -ve s 5 fon 2.so olv = -1.5omis sign indicates is flravelling to the left At = .06255. DP _-2.50 (7.50-(-250) ot 0625 2.50x 12.50 0625 OP 4.90( 488-(-150) SOON. DT .0625 N = - 500 Fon 4.90kg Applying momentum conservation prhiple 8x28 ² = 3 x 50 (9) + 2x40 i tzv wt S Y Am q Х > 30 = 8x28 ? - 80 - 1soj - 3 = 1442 - 150 Î 48î - 5o 3 m/s 3 After collisia 4 m=180kg V = 31 m/s 475kg m = 475kg m. L 350 S Before collision 670 V - 18 m/s Loum wate Applying momentum conservation 475 ( 31cos 35° ſ +31 Sin3s®3) 475 475 18 cos 67° + 185ın 67°4) + 180 To V = 48.45 ŷ + 3. 2 A Aug I IVI = 48.56 m/s 86.220 Ajanta Oxford PRODUCTS