A 1400 sedan goes through a wide intersection traveling from north to south when

Question

A 1400 sedan goes through a wide intersection traveling from north to south when it is hit by a 2400 SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.60 west and 6.16 south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Explanation / Answer

   Mass of sedan   m   =   1400   kg    mass of SUV   M   =   2400   kg   Let the south lie over x axis such that the unit vector along south is i^ and unit vector towards west is j^ ( taking east to west along y axis).   initial velocity of sedan   u   =   u i^    initial velocity of SUV   U   =   U j^    Final velocity of enmeshed mass (m + M)   =   v    final displacement   d   =   (6.16 i^   +   5.60 j^)   m      magnitude of displacement    |d|   =   v(6.162   +   5.602)                   =   8.325   m    Force of friction   F   =   µk * (m + M) * g                                     =   0.750 * (1400 + 2400) * 9.8                                     =   2.793 * 104   N    Work done against friction   W   =   F * d                                                        =   2.793 * 104 * 8.325                                                    =   2.325 * 105   J    Now this must be equal to k.e. of both vehicles, i.e.    (1/2) * m * u2   +   (1/2) * M * U2   =   W    0.5 * 1400 * u2   +   0.5 * 2400 * U2   =   2.325 * 105    7 * u2   +   12 * U2   =   2325           ------------ (1)       also       (1/2) * m * u2   +   (1/2) * M * U2   =   (1/2) * (m + M) * v2             =>   0.5 * (1400 + 2400) * v2   =   2.325 * 105    v2   =   2.325 * 105 / 1900    speed of enmeshed vehicals just after collision    v   =   v122.368   =   11.06   m/s From law of conservation of momentum    m * u   +   m * U   =   (m + M) * v    1400 * u   +   2400 * U   =   3800 * 11.06    7 * u   +   12 * U   =   210.14    U   =   (210.14 - 7 * u) / 12   Substituting this value in equation (1)    7 * u2   +   12 * {(210.14 - 7 * u) / 12}2   =   2325    84 * u2   +   (210.14   -   7 * u)2   =   2.79 * 104    84 * u2   +   4.416 * 104   +   49 * u2   -   2.94196 * 103 * u   =   2.790 * 104    133 * u2   -   2941.96 * u   +   16260   =   0    solving this quadratic equation    u   =   10.80   m/s      or      u   =   11.32   m/s    then   U   =    (210.14 - 7 * u) / 12    U   =   16.61   m/s   or      U   =   10.91   m/s   or              133 * u2   -   2941.96 * u   +   16260   =   0    solving this quadratic equation    u   =   10.80   m/s      or      u   =   11.32   m/s    then   U   =    (210.14 - 7 * u) / 12    U   =   16.61   m/s   or      U   =   10.91   m/s   or              

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