4. [2pt] A drunken driver crashes his sports car (mass m ) into a stationary sta

Question

4. [2pt]
A drunken driver crashes his sports car (mass m) into a stationary station wagon (mass M), that was stopped at a red traffic light. The cars stuck together and slid with locked wheels for a distance D before coming to rest. The coefficient of sliding friction on the dry road was mu. What was the speed of the sports car before it hit the station wagon?

Enter a symbolic answer as a function of m, M, D, mu, and g. You may also use numbers such as 3.0 or 0.5. As with all symbolic answers, do not ever include units or unit conversion factors. For example, you might enter:

Explanation / Answer

Applying law of conservation of momentum

mv1i+Mv2i = m1v1f+Mv2f

but v2i =0 and v1f =v2f = V

m1v1i= (m+M)V ---------------------(1)

V is final velocity of two body together after the collision,

Now when two body stuck together move with combine mass= m+M they have friction coefficient =mu

By Newton’s second law

Ff = ma

mu*(m+M)g = ma

thus acceleration of combine mass after collision = a= mu*g

Now applying kinematic equation,

vf^2=vi^2+2a*d

we will find relation between v , mu and D

V^2 = 0^2 +2*mu*g*D

V= sqrt(2*mu*g*D) -----------------(2)

Plugging (2) in (1)

v1i = [(m+m)/m]* sqrt(2*mu*g*D)

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