**********************PLEASE ANSWER IT STEP BY STEP AND EXPLAIN REALLY WELL I AM

Question

**********************PLEASE ANSWER IT STEP BY STEP AND EXPLAIN REALLY WELL I AM A SLOW LEARNER******************

An automobile of mass M1= 1500 kg traveling east at speed V1=12.5 m/s collides with a second automobile of mass M2= 1800 kg that is moving with speed V2= 10.0 m/s in a direction that is (theta) 36.9(degree) north of west . the collision is tottaly inelastic so that the two automobiles move together after the collision.

A) keeping in mind that momentum is a vector , find the x and y componets of the total momentum of the system before the collision ( let the x-axis point east and the y-axis point north)

B) using the result in part A, find the magnitude of the total momentum of the system and its direction. For direction, especify the angle that the total mementum vector makes with the positive x-axis.

C) Using momentum convervation, calculate the speed of the two automobiles ( moving as one) after the collision.

D) find the kinetic energy that is lost at the collision.

Explanation / Answer

A) before collision,

V1 = 12.5i

V2 = 10(-cos36.9i + sin36.9j) = - 8i + 6j m/s

momentum = M1V1 + M2V2

pi = 1500(12.5i) + 1800 ( - 8i + 6j)   = 4350i + 10800j kg m/s

pi_x = 4350 kg m /s

pi_y = 10800 kg m/s

2) magnitude = sqrt(pi_x^2 + pi_y^2 ) = 11643.13 kg m/s

direction = tan^-1(pi_y / pi_x) = 68.06 north of east.

3) using moemntum conservation,

pi = pf

4350i + 10800j = (1500 + 1800)v

v = 1.32i + 3.27j m/s

speed = sqrt(1.32^2 + 3.27^2) = 3.53 m/s

direction same as momentum

4) initial KE, ki = 1500 x 12.5^2 /2 + 1800 x 10^2 /2 = 207187.5 J

Kf = (1500 + 1800) 3.53^2 /2 = 20560.48 J

energy lost = Ki - Kf = 186627.02 J

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