Physics- Velocity and Kinetic Energy?

Two particles of mass m1 = 1.7 kg and m2 = 3.5 kg undergo a one-dimensional head-on collision as shown in the figure below. Their initial velocities along x are v1i = 15 m/s and v2i = - 7.5 m/s. The two particles stick together after the collision (a completely inelastic collision). (Assume to the right as the positive direction.)

a) Find the velocity after the collision. (m/s)

b) How much kinetic energy is lost in the collision? (J)

3 Answers

  • Anonymous
    9 years ago
    Favorite Answer



    since they the collision is inelastic


    making the equation


    (1.7kg)(1.5m/s)+(3.5kg)(-7.5m/s)=(1.7kg + 3.5kg)vf



    ***vf=-4.56 m/s***

    For the kinetic energy

    is the KE (lost) for each particle?

    if I got that right,

    use the formula:

    Just substitute values and use

    KE=(1/2)m1v1f-m1v1i----->for KE lost in particle 1


    KE=(1/2)m2v2f-m2v2i----->for KE lost in particle 2

    Hope this helps you!


  • baden
    Lv 4
    4 years ago

    properly quite lots you have already got the different men explaining to you what to do with this issue. properly in certainty conservation of potential, what does it mean? properly conservation potential not something is lost so for you to use potential potential and convert it into Kinetic potential PE= mgh KE= a million/2mv^2 so first you calculate the potential potential first so utilising the formuela given above: PE= 100X9.81X50 = 49050J and we pronounced in the previous that no potential is lost so we are in a position to transform it into the Kinetic potential 49050= a million/2X 100X v^2 so make v the region and you get the respond 31.3209... rounding it to the smallest sig we get 30J

  • 9 years ago

    do it by law of conservation of mass

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