Please help a momentum problem?

Comments

• liberty

  June 2021

  Here's the link for conservation of momentum.

  http://en.wikipedia.org/wiki/Momentum

  m1*u1 + m2*u2 = m1*V1 + m2*V2

  Where,

  m1 = mass of cart 1 = 737 g = 0.737 Kg

  m2 = mass of cart 2 = 867 g = 0.867 Kg

  u1 = velocity of cart 1 before collision = 6.26 m/s

  u2 = velocity of cart 2 before collision = 0 m/s

  V1 = velocity of cart 1 after collision

  V2 = velocity of cart 2 after collision

  After the collision cart 1 will move at a V1 from the following equation:

  V1 = [ (m1-m2) / (m1+ m2) ]*u1 + [ (2*m2) / (m1+m2) ]*u2

  and cart 2 will move at a V2 from the following equation:

  V2 = [ (m2-m1) / (m1+ m2) ]*u2 + [ (2*m1) / (m1+m2) ]*u1

  We then have:

  V1 = [(737-867)/(737+867)] * 6.26 m/s + 0 = - 0.51 m/s

  Notice: negative sign on V1 means that cart 1 move backward after the collision.

  V2 = 0 + [2(737) / (737+867)] * 6.26 m/s = 5.75 m/s

  Notice: you can solve this problem using mass unit in either g or Kg and still get the same answers.
• letx27sdaretoansweranything

  June 2021

  center of mass velocity = {(737*6.26)/(737+867)} = 2.88 m/s

  Before collision-

  In center of mass frame: u1 = 6.26 - 2.88 = 3.38 m/s; and u2 = 0-2.88 = -2.88 m/s

  After collision-

  In center of mass frame: v1 = -3.38 m/s; and v2 = -(-2.88) = 2.88 m/s

  After collision-

  In laboratory frame: V1 = -3.38+2.88 = -0.50 m/s; and V2 = 2.88 +2.88 = 5.76 m/s

  So cart1 will recoil with 0.50 m/s and cart2 will move forward with 5.76 m/s