Question

# Car #1 with a mass of 1.50 x 103 kg is travelling east at a speed...

Car #1 with a mass of 1.50 x 103 kg is travelling east at a speed of 25.0 m/s. It collides in the middle of an intersection with Car #2 which has a mass of 2.50 x 103 kg and enters the intersection travelling north at a speed of 20.0 m/s. (a) Find the magnitude and direction of the velocity of the wreckage, assuming that after the collision the two cars stick together and that frictional forces can be neglected. (b) How much kinetic energy is lost in the collision?

Given,

m1 = 1500 kg ; u1 = 25 m/s ;

m2 = 2500 kg ; u2 = 20 m/s

intitial momentum along x and y is:

Pix = 1500 x 25 = 3.75 x 10^4 kg-m/s

Piy = 2500 x 20 = 5 x 10^4 kg-m/s

from conservation of momentum

(1500 + 2500)vf cos(theta) = 3.75 x 10^4 kg-m/s

(1500 + 2500)vf sin(theta) = 5 x 10^4 kg-m/s

dividing 2 eqn by 1

theta = tan^-1(5/3.75) = 53.13 deg

Now, from 1st

vf = 3.75 x 10^4/(1500 + 2500) cos(theta)

vf = 3.75 x 10^4/4000 x cos53.13 = 15.62 m/s

Hence, vf = 15.62 m/s ; theta = 53.13 deg

b)KEi = 1/2 m1 u1^2 + 1/2 m2 u2^2

KEi = 0.5 (1500 x 25^2 + 2500 x 20^2) =968750 J

KEf = 0.5 x (1500 + 2500) x 15.62^2 = 487968.8 J

KE(lost) = 487968.8 - 968750 = -480781.2 J

Hence, KE(lost) = -480781.2 J = 4.81 x 10^5 J

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