Original Problem: A 2500 kg car traveling 18 m/s collides witha 5700 kg truck tr

Question

Original Problem: A 2500 kg car traveling 18 m/s collides witha 5700 kg truck traveling 10 m/s and they stick together. Originally the cars were traveling at right angles. What isthe speed and direction of the two vehicles after thecollision? I tried applying the formula: m1v1+ m2v2= (m1 +m2)vc This formuala provided the answerof vc= 12.44 m/s. Unfortunately this is not the answer proved in the key from theprof. I am certain the right angle has something to do withit but I am not sure how to appply that theory. So myquestion is how do I solve for the speed after the collision andhow do I determine the direction. (all the sample problemsinvolve objects in a single line diverting to differentdirections!) Original Problem: A 2500 kg car traveling 18 m/s collides witha 5700 kg truck traveling 10 m/s and they stick together. Originally the cars were traveling at right angles. What isthe speed and direction of the two vehicles after thecollision? I tried applying the formula: m1v1+ m2v2= (m1 +m2)vc This formuala provided the answerof vc= 12.44 m/s. Unfortunately this is not the answer proved in the key from theprof. I am certain the right angle has something to do withit but I am not sure how to appply that theory. So myquestion is how do I solve for the speed after the collision andhow do I determine the direction. (all the sample problemsinvolve objects in a single line diverting to differentdirections!)

Explanation / Answer

Let         ma = 2500 kg   ; va = 18m /s         mb = 5700 kg   ; vb  = 10 m /s As they are at right angles to each other .             vx   = ( ma   va  ) /(   ma +   mb )                   =   ( 2500 * 18) / ( 2500 + 5700)                   = 5.49 m /s              vy  = ( mb   vb  ) /(   ma +   mb )                    =   ( 5700* 10) / ( 2500 + 5700)                    = 6.95 m /s Velocity after collision is :                      V = (vx)2 +(vy)2                             =  (5.49)2 +(6.95)2                               = 8.577 m /s and direction is :               = tan-1( vy  / vx )                  = tan-1( 6.95  / 5.49 )           = 51.69 Hope this helps u!                    =   ( 5700* 10) / ( 2500 + 5700)                    = 6.95 m /s Velocity after collision is :                      V = (vx)2 +(vy)2                             =  (5.49)2 +(6.95)2                               = 8.577 m /s and direction is :               = tan-1( vy  / vx )                  = tan-1( 6.95  / 5.49 )           = 51.69 Hope this helps u!

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.