Question

The one direction tour van (2100 kg total) drives at 23 m/s to the right and collides with a 950 kg parked car on level ground. The car is neutral and doesn't have its parking brakes engaged. The van doesn't apply its brakes during the collision. The collision takes 0.17 s. Just after the collision, the car is observed to be moving 18 m/s to the right.

a) find the velocity of the van after the collision

b) impulse on the van

c) impulse on the car

d) max force on the car by the van

e) max acceleration of the car

f) max acceleration of the van

g) if the collision time was decreased, how would items a-f change?

Answer 1

Matt 4 van, M= 2000 kg Most of can, m=g5oky Shital speed q Yen, U = 232, rutal geed y Can Hyom/ Afley could speed o can; V2 - 18 M/S t velou ry 9 van after collision be v fi Must muy = MVI + MY2 2100X23+ 0 = 2100 V + 950X18 V - 14.86 m . in the same direbon as . the initial direchon Towards right V = 14-86 mle sight! Impulse on van, Be IV - Aprop 2100 (14.86-23) NL. IV = -17100 N-S 5 Iv - 17100 N-4 left towards let © Impulse on car, Ic - Akar 960(16-0) NS 1 Tc=17100 Nes right trucurdy right CD Max dorze, ťa 17.000 N On Card PE 100588 N +Right @ Max accele vashion, Ames, e o s 100584 mis? 9 the car 950 a max, & 10509 m/² right] © 2100 Max acele schon qe the van. Qoux, = F = -100585 m i amar, -47.9 m/? ..Jamur, v = 4709 mye? left Is the collition time to decreased then answer 7 part (a), (b) c) will remain unchange (same as before) white and wer of part , Ô will change Answer q d, e, it will increase in value 9