I WILL give a Medal for the best response! A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. After the collision, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? The correct answer is supposed to be 458 J. How can I solve this problem? I already tried using conservation of momentum, and conservation of energy, but I'm not getting the right answer. Am I supposed to be using calculus to solve this problem? Any help would be greatly appreciated!
Hi
I believe I have your solution. Still care for it?
It's okay. I was able to solve it with help from someone in the Physics study area. Thank you though!
I gave you a medal anyway because you were the only one to answer! :)
Lol, sorry, I have your solution...one sec.
Mass of the firstbody, M1 = 2 kg Mass of the secondbody, M2 = 3 kg Velocity of thefirst body before collision, U1 = 20 m/s Velocity of thesecond body before collision, U2 = - 10 m/s ( '-' representWest ) Velocity of thefirst body after collision, V1 = - 5 m/s Velocity of thesecond body after collision = V2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ From the lawof conservation of momentum, Totalmomentum after collision = Total momentum before collision M1 V1 + M2 V2 = M1 U1 + M2 U2 2 * - 5 + 3 V2 = 2 * 20 + 3 * - 10 - 10 + 3 V2 = 40 - 30 - 10 + 3 V2 = 10 3 V2 = 20 Velocity of the second body after collision, V2 = 6.67 m/s ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Total kinetic energy before collision, E1 = ( 1/2 ) M1 U12 + ( 1/2 ) M2 U2 2 = 0.5 [ 2 * 20 2 + 3 * ( -10 ) 2 ] = 0.5 * 1100 = 550 J Total kinetic energyafter collision, E2 = ( 1/2 ) M1 V1 2 + ( 1/2 ) M2V2 2 = 0.5 [ 2 * ( - 5 ) 2 + 3 * ( 6.67 )2 ] = 0.5 * 183.47 = 91.73 J Loss of kinetic energy during the collision, ΔE = E1 - E2 = 550 - 91.73 = 458.27 J Anyways, hope that made sense. Good luck!
This is very good! Thank you!
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