Question

A 10 ton freight car (call it car A) is moving at 4 m/s collided with a 30 ton freight car (call it car B), initially at rest. After the collision, they are stuck together. Assume the rolling friction of the cars is negligible

final momenta? 6. (18 points) A 10 ton freight car (call it car A) moving at 4 m/s collides with a 30 ton freight car (Can initially at rest. After the collision, they are stuck together. Assume the rolling friction of the cars is negligible, 30 Tons 4 m/s 0 m/s Before 7 m/s After (a) (3 points Use momentum conservation to find the final velocity of the two cars after the collision (b) (3 points) How much work did car A do on car B? Be sure to use the proper sign (+/-). (c) (3 points) How much work did car B do on car A? Be sure to use the proper sign (+/-). (d) (3 points) How much did the total kinetic energy of the system change? What happened to this energy? le 13 points) Why were you able to solve this problem using momentum conservation, but not energy conser- vation? 13 points) Now say car B, instead of initially being at rest, is initially moving to the left with a speed of 2 m/s The 10 ton car is still moving to the right at 4 m/s. Find the final speed of the two cars in this scenario. in tan train is moving to the right along the tracks at 7 m/s It Anmaches from hehind a 18

please help woth number 6. please show all work and all of the equations used.

Answer 1

slution. NAB. Veum vero A malo m= Bo m=40 ton mi m milo tou initial momentum of two care. momentum Pi = MAVAT MB NB. = 10x4 + 3000 pi= 40 ton-mis cars. final momentum of two P = MARX VAB - 40 VAB- Die = Pf 40. VAB = 40 I VAB = 40 = 1 ms] ③ work done on carß = change is kinetic energy of cars. 1. mn. As Ilon= 1000kg = 1 * (30 ton) .1 1x 30 x1000 = a 15000 joules e o work done on car A by carb = change is kee of cara = 13. MA (UAVA 1x 10x1000 (12 - 42 ) = – 45 000 poules @ Initial kot of system = 1 mA. VA = 4 x 10x1000 x 42 - Bo, ooojaules. final ke of syten 1. MAB. = 2 40,000 x 2 = 20,000 Joules OKE = – 60,000 joules, this energy is lost is collision

slution. NAB. Veum vero A malo m= Bo m=40 ton mi m milo tou initial momentum of two care. momentum Pi = MAVAT MB NB. = 10x4 + 3000 pi= 40 ton-mis cars. final momentum of two P = MARX VAB - 40 VAB- Die = Pf 40. VAB = 40 I VAB = 40 = 1 ms] ③ work done on carß = change is kinetic energy of cars. 1. mn. As Ilon= 1000kg = 1 * (30 ton) .1 1x 30 x1000 = a 15000 joules e o work done on car A by carb = change is kee of cara = 13. MA (UAVA 1x 10x1000 (12 - 42 ) = – 45 000 poules @ Initial kot of system = 1 mA. VA = 4 x 10x1000 x 42 - Bo, ooojaules. final ke of syten 1. MAB. = 2 40,000 x 2 = 20,000 Joules OKE = – 60,000 joules, this energy is lost is collision

slution. NAB. Veum vero A malo m= Bo m=40 ton mi m milo tou initial momentum of two care. momentum Pi = MAVAT MB NB. = 10x4 + 3000 pi= 40 ton-mis cars. final momentum of two P = MARX VAB - 40 VAB- Die = Pf 40. VAB = 40 I VAB = 40 = 1 ms] ③ work done on carß = change is kinetic energy of cars. 1. mn. As Ilon= 1000kg = 1 * (30 ton) .1 1x 30 x1000 = a 15000 joules e o work done on car A by carb = change is kee of cara = 13. MA (UAVA 1x 10x1000 (12 - 42 ) = – 45 000 poules @ Initial kot of system = 1 mA. VA = 4 x 10x1000 x 42 - Bo, ooojaules. final ke of syten 1. MAB. = 2 40,000 x 2 = 20,000 Joules OKE = – 60,000 joules, this energy is lost is collision

Borth cars move rogether of her collision. v=ums v=amus , AB VAB meloton MAB=40ton. "=30ron we=30ton Here again momentum will remain conserved. = MABV MAINA & MB VB BV lox 4 + 30x2 40 (VAR 2.5 mls]

Borth cars move rogether of her collision. v=ums v=amus , AB VAB meloton MAB=40ton. "=30ron we=30ton Here again momentum will remain conserved. = MABV MAINA & MB VB BV lox 4 + 30x2 40 (VAR 2.5 mls]

Explanation for (e) part is that in an inelastic collision energy is not conseved.