A car of mass 2200 kg collides with a truck of mass 4400 kg at the intersection

Question

A car of mass 2200 kg collides with a truck of mass 4400 kg at the intersection of horizontal roads. The vehicles stick together and slide immediately after the collision. The x- and y-components of the car’s velocity just before the collision are (41, 0) m/s, respectively, and the same components of the truck’s velocity just before the collision are (-19, 28) m/s. -Than you!

Calculate the x and y-component of the velocity of the combined car and truck immediately after the collision, in meters per second.

This collision is highly inelastic. Calculate the loss of total kinetic energy during the collision, in joules.

Find the increase of the internal energy of the car and truck (due to heating and deformation), in joules.

Explanation / Answer

Ans:-

For all collisions where no external forces are involved, the linear momentum

p=mv

is conserved.

Let the car move along the x axis before the collision. Therefore truck moves along the x and y axis before the collision. Initial momentum is

Pinitial = mcar*vcar + mtruck*vtruck
=2200*41 x + 4400*(-19 x + 28 y)

From x component
2200*41 – 4400* 19 =6600*vfinal
vfinal=1m/s

From y component
4400*28=6600*vfinal

Vfinal = 18.67m/s

--.-.-.-.-.-.-..-.-.-.-.-.-.-.-.-..-.-.-.-.-.-.-.-.-.-
(b) Kinetic energy before collision

KEbefore=KEcar+KEtruck
=1/2*mcar*v^2car+1/2*mtruck*v^2truck
=1/2*2200*41^2 x +1/2*4400*(-19)^2 x + ½ 4400*28^2y
=1849100 x +794200x +1724800y

=2643300x+ 1724800y J

=3156258.85J…………………………take magnitude

Kinetic energy after collision

KEafter=1/2*(mcar+mtruck)*v^2final

=1/2*6600*(1^2x +18.67^2y)

=3300*(1x+348.57y)

=1153582.09J ………….take magnitude

Now

KElost=KbeforeKEafter=3156258.851153582.09=2002676.76J

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