(hrwc10p64_6e) A barge with mass 1.3×10 5 kg is proceeding down river at 6.1 m/s

Question

(hrwc10p64_6e) A barge with mass 1.3×105kg is proceeding down river at 6.1 m/s in heavyfog when it collides broadside with a barge heading directly acrossthe river (see the figure). The second barge has mass2.58×105kg and was moving at 4.2m/s. Immediately after impact, the second barge finds itscourse deflected by 18° in the downriver direction and itsspeed increased to 5.0 m/s. The river current waspractically zero at the time of the accident. What are the speedand direction of motion of the first barge immediately after thecollision? Speed? Enter the angle of deflection from the originaldirection with + as motion to its left and - as to itsright How much kinetic energy is lost in thecollision? Enter the angle of deflection from the originaldirection with + as motion to its left and - as to itsright How much kinetic energy is lost in thecollision?

Explanation / Answer

Consider the first barge to be moving in the -y direction, and thesecond to be moving in the +x direction. Conserve the momentum in both directions. M1V1i+M2V2i=M1V1f+M2V2f (1.3x105)(6.1)(-ydirection)+(2.58x105)(4.2)(xdirection)=(1.3*105)(Vx)+(1.3*105)(Vy)+(2.58*105)(5.0*sin(18))(-ydirection)+(2.58*105)(5.0*cos(18))(x direction) Now separate the variables by direction. (1.3*105)(6.1)=(1.3*105)(Vy)+(2.58*105)(5*sin(18)) Vy=3.0336m/s (2.58*105)(4.2)=(1.3*105)(Vx)+(2.58*105)(5*cos(18)) Vx=-1.10 (so - x direction) So total V=(3.03362+(-1.10)2)=3.227m/s Direction = tan-1(3.0336/-1.10) = -70 deg. from the xaxis, so since it was initially travelling in the y, it's deflectedby (90+-70), or 20 degrees. The + or - requires the image, but it'sthe opposite direction of the other barge. As for the loss of KE 1/2(M1V12)+1/2(M2V22)=Elost+1/2(M1V12)+1/2(M2V22) Substituting in the values and solving for E, we find that thesystem loses 795264J of KE.

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