A 2000-kg car traveling at 10.0 m/s along positive x-direction collides with a 3

Question

A 2000-kg car traveling at 10.0 m/s along positive x-direction collides with a 3000-kg car moving with an initial velocity of 8.0m/s @ 120 degree. The cars stick together and move 5.00 m before friction causes them to stop. What is the speed of the two cars right after collision? Determine the magnitude and direction of the impulse on each car. You can skip one set of calculations if you can clearly justify your answer. Determine the coefficient of kinetic friction between the cars and the road, assuming that the acceleration was constant.

Explanation / Answer

here,

mass of car 1 , m1 = 2000 kg

mass of car 2 , m2 = 3000 kg

speed of car 1 , u1 = 10 m/s i

speed of car 2 , u2 = 8 * ( cos(120) i + sin(120) j)

u2 = -4 m/s i + 6.93 m/s j

a)

let the final speed be v

using conservation of momentum

m1*u1 + m2 * u2 = ( m1 + m2) * v

2000 * 10 i + 3000 * (-4 m/s i + 6.93 m/s ) = 5000 * v

v = 1.6 m/s i + 4.16 m/s j

|v| = 4.46 m/s

the speed of car right after the collison is 4.46 m/s

b)

for car 1

impulse , I = m1 * ( v - u1)

I = 2000 * ( 1.6 m/s i + 4.16 m/s j - 10 m/si)

I = ( - 16800 kg.m/s i + 8000 kg.m/s j)

for car 2

impulse , I = m2 * ( v - u2)

I = 3000 * ( 1.6 m/s i + 4.16 m/s j + 4 m/s i - 6.93 m/s j )

I = ( 16800 kg.m/s - 8310 kg.m/s j)

c)

distance travelled , d = 5 m

let the coefficient of friction be uk

accelration due to friction a = - uk*g

using third equation of motion

v^2 - u^2 = 2 * a*d

0 - 4.46^2 = 2*uk*9.8*5

uk = 0.203

the coefficient of friction is 0.203

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