Question

A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 14 m/sm/s collides with the rear of the stalled car. After being hit, the 750 kg car slides on the ice at 4 m/sm/s in a direction 30∘ north of east.

Part A

What is the magnitude of the velocity of the 1000 kg car after the collision?

Express your answer in meters per seconds to two significant figures.

Part B

What is the direction of the velocity of the 1000 kg car after the collision?

Express your answer in degrees to two significant figures. Enter a positive value if the direction is north of east and negative value if the direction is south of east.

Part C

Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

Express your answer to two significant figures.

Answer #1

here,

the mass of car , m1 = 750 kg

the mass of second car ,m2 = 1000 kg

the initial velocity of second car , u2 = 14 i m/s

the final velocity of first car , v1 = 4 m/s * ( cos(30) i + sin(30) j)

v1 = (3.46 i m/s + 2 j m/s)

let the final velocity of second car be v2

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

750 * 0 + 1000 * ( 14 i) = 750 * (3.46 i + 2 j) + 1000 * v2

v2 = ( 11.4 i m/s - 1.5 j m/s)

the magnitude of final velocity of 1000 kg car , |v2| = sqrt(11.4^2 + 1.5^2) m/s

|v2| = 12 m/s

b)

the direction of final velocity of 1000 kg car , phi = arctan(1.5/11.4)

phi = 7.5 degree South of East

c)

the ratio of the kinetic energy of the two cars just after the collision to that just before the collision , R = KEf /KEi

R = (0.5 * m1 * v1^2 + 0.5 * m2 * v2^2) /(0.5 * m1 * u1^2 + 0.5 * m2 * u2^2)

R = (750 * 4^2 + 1000 * 14.5^2) /(1000 * 14^2)

R = 1.1

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