Question

Q5. Part a) The 5500 kg bus B is traveling to the right at 20 m/s. Meanwhile a 1100 kg car A is traveling at 15 m/s to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision. Hint: use principles of impulse and momentum. Q5. Part b) Calculate kinetic energy of the bus before impact. 05. Part c) Calculate kinetic energy of the car before impact. Q5. Part d) Calculate kinetic energy of the bus+car (tangled together) immediately after collision 03 = 20 m/s B - 15 m/s Value Unit Write your final answers in the table below. Question Parameter 1a Vous car 1b Thus 1c Tear 1d Tbuscar le Elost

Answer 1

Detailed solution is as shown in the attached images.

The results obtained are :

a) 19.166 m/s

b) 1100 KJ

c) 123.75 KJ

d) 1212.207 KJ

e) 11.543 KJ

solution : Let V and Va be initial velocities and bus | car respectively. Let UAB the be relocity common Given VA = 15 ml MA 11Do 4 = 5500 kg VB = 20mle B UB VA VAIB a) Conservation of linear momentum, m uit M₂ V2 = m, 0,' + mq ,' Here, MAVA +MBUB =(MA AMB] VAR 1100 (15) + 5500 (20) (1100+ +550o) VAIB NAB 19. 166 ms This is the common velocity after collision. b) Kinetic Energy of bus before impact ราช 1 / 2 (5.500] (208² 1100 KJ - KE of bus warts 100 kilo joules

c) KE of car before impact Žmv2 L (1100) (15)² 123.75 Kilo Toules d) KE of tangled bus & car 2 = È (MAIMA) VA = = (1100 +5500) (19.766)2 1212207 KJ) e) Energy lost due to collision initial final [1100 + 123.75 75] - [1212.207] 11.548 KJ Flost 11.54 kТ -X