A 15500 N automobile travels at aspeed of 46 km/h northward along astreet, and a

Question

A 15500 N automobile travels at aspeed of 46 km/h northward along astreet, and a 8500 N sports cartravels at a speed of 56 km/h eastwardalong an intersecting street. (a) If neither driver brakes and the carscollide at the intersection and lock bumpers, what will thevelocity of the cars be immediately after the collision?
magnitude 1 km/h direction 2° (counterclockwise fromdue east)
(b) What percentage of the initial kinetic energy will be lost inthe collision?
3% (a) If neither driver brakes and the carscollide at the intersection and lock bumpers, what will thevelocity of the cars be immediately after the collision?
magnitude 1 km/h direction 2° (counterclockwise fromdue east)
(b) What percentage of the initial kinetic energy will be lost inthe collision?
3% magnitude 1 km/h direction 2° (counterclockwise fromdue east)

Explanation / Answer

weight of auto mobile W = 15500 N mass of auto mobile M = W / g                                    = 1581.632 kg weight of car w = 8500 N mass of car m = w / g                       = 867.34 kg let the unit vector in east direction be i  and in north direction be j   then initial speed of automobile U = 46 j initial speed of car u = 56 i let the velocity of the cars after collision be v and it makes an angle with east direction . In north direction : ------------------ apply law of conservation of linear momentum , MU + 0 =( m + M ) v sin                                                                        v sin = MU / ( m + M )                                                                                    = 29.7        -----( 1) In east direction : ------------------ apply law of conservation of linear momentum ,  mu +0 = ( m + M ) v  cos                                                                        v cos = mu / ( m + M )                                                                                    = 19.83         -----(2) eq ( 1 ) / eq ( 2 ) ==>   tan = 29.7 /19.83                                           = 56.27 degrees substitue this in eq ( 1 ) we get   v sin 56.27 = 29.7 from this speed of the cars after collision v = 35.71 m/ s In east direction : ------------------ apply law of conservation of linear momentum ,  mu +0 = ( m + M ) v  cos                                                                        v cos = mu / ( m + M )                                                                                    = 19.83         -----(2) eq ( 1 ) / eq ( 2 ) ==>   tan = 29.7 /19.83                                           = 56.27 degrees substitue this in eq ( 1 ) we get   v sin 56.27 = 29.7 from this speed of the cars after collision v = 35.71 m/ s

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