A 1, 700 kg van moving with a speed of 4 m/s collides head-on with a 1, 200 car
ID: 1524121 • Letter: A
Question
A 1, 700 kg van moving with a speed of 4 m/s collides head-on with a 1, 200 car that is moving at a speed of 2 m/s in the opposite direction. After the collision, the two cars lock together and the wreckage slides to a stop. The coefficient of friction (mu) between the wreckage and the road is a. What is the speed of the wreckage immediately after the collision? b. What is the acceleration (deceleration) of the wreckage after the collision? c. How long will it take the wreckage to slide to a stop after the collision is over? d. How far will the wreckage travel after the collision before it finally stops? e. 0.1 seconds before the cars come to a stop they hit a blockade in the middle of the road. If they undergo a constant deceleration (constant force of 14, 137 N) by the blockade until they come to a stop how long will it take for the wall to stop them? f. In words explain why the cars can lose momentum without the wall appearing to move. Why does this not violate the law of conservation of momentum?Explanation / Answer
a) conservation of momentum
m1u1+m2u2 = (m1+m2)v
v = 1700*4-1200*2/2900
v = 1.52 m/s
b) f = ma
mue*mg = ma
a = -mue*g = -0.4*9.8 = -3.92 m/s^2
c) v = u+at
t = 1.52/3.92 = 0.387 sec
d) from v^2-u^2 = 2as
s = 1.52^2/2*3.92 = 0.295 m
e) from impulse momentum
f*t = mv
t = 2900*1.52/14137 = 0.312 sec
momentum is always conserved in all types of collision but energy is not conserved in an inelastic collisions
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.