Becasue of brake failure an automobile parked on a hill of slope 1:10 rolls 12 m

Question

Becasue of brake failure an automobile parked on a hill of slope 1:10 rolls 12 m downhill and strikes a parked automobile. The mass of the first automobile is 1400 kg and the mass of the second automobile is 800 kg. Assume that the first auto rolls without friction and the collision is elastic.

(a) What are the velocities of both automobiles immediatily after the collision?

(b) After the collision, the first automobile continues to roll downhill, with acceleration, and the second automobile skids downhill, with deceleration. Assume that the second auto skids with its wheels locked, with a coefficient of sliding friction of 0.90. At what time after the first collision will the automoiles have another collision, and how far from the initial collision?

Explanation / Answer

slope = 0.1

theta = tan^-1(0.1)= 5.71 degrees

let h is the height

h = L*sin(5.71) = 12*sin(5.71) = 1.194 m

m1 = 1400 kg

m2 = 800 kg

let u1 is the spped os first automobile beforre collision.

0.5*m1*u1^2 = m1*g*h

u1 = sqrt(2*g*h) = sqrt(2*9.8*1.194) = 4.838 m/s

before collision, the velsocity of secodn body, u2 = 0
a)

a) let v1 and v2 are the velocties of auto mobiles after the collision.

v1 = (m1-m2)*u1/(m1+m2) + 2*m2*u2/(m1+m2) = (1400-800)*4.838/(1400+800) = 1.319 m/s



v2 = (m2-m1)*u2/(m1+m2) + 2*m1*u1/(m1+m2) = 0 + 2*1400*4.838/(1400+800) = 6.157 m/s


b)

aceleration of first automobile, a1 = g*sin(5.71) = 0.975 m/s^2

deceleration of first automobile, a1 = g*sin(5.71)-g*cos(5.71)*mue = -7.8 m/s^2


let t is the time taken for another collision.


diatnce travelled by tow automobiles is same.

d1 = d2

v1*t + 0.5*a1*t^2 = v2*t + 0.5*a2*t^2

1.319*t + 0.5*0.975*t^2 = 6.157*t - 0.5*7.8*t^2
4.838*t = 0.5*t^2*(7.8+0.975)

4.838 = 0.5*7.605*t

==> t = 4.838*2/7.605 = 1.272 s

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