Question #102110

2.Inelastic Collision. Two automobiles of 540 and 1400 kg collide head-on while


moving at 80 km/h in opposite directions. After the collision the automobiles remain locked together.


(a)Find the velocity of the wreck immediately after the collision.


(b)Find the kinetic energy of the two-automobile system before and after the collision.


(c)The front end of each automobile crumples by 0.60 m during the collision. Find the acceleration (relative to the ground) of the passenger compartment of each automobile;

make the assumption that these accelerations are constant during the collision.

Expert's answer

a) From the conservation of momentum:


−mv+Mv=(M−m)v=(m+M)u-mv+Mv=(M-m)v=(m+M)u

u=803.61400−5401400+540=9.85msu=\frac{80}{3.6}\frac{1400-540}{1400+540}=9.85\frac{m}{s}

b)


Ki=0.5(mv2+Mv2)=0.5(m+M)v2K_i=0.5(mv^2+Mv^2)=0.5(m+M)v^2

Ki=0.5(540+1400)8023.62=479000 JK_i=0.5(540+1400)\frac{80^2}{3.6^2}=479000\ J

Kf=0.5(540+1400)9.852=94000 JK_f=0.5(540+1400)9.85^2=94000\ J

c)


u2=2ad→9.852=2(0.6)au^2=2ad\to 9.85^2=2(0.6)a

a=81ms2a=81\frac{m}{s^2}


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