A 2.0 kilogram cart moving due east at 6.0 meters per second collides with a 3.0 kilogram cart moving due west. the carts stick together and come to rest after the collision. what was the initial speed of the 3.0 kilogram cart?
The total momentum must conserve before and after the collision.
1) Calling [tex]m_1=2~Kg[/tex] and [tex]m_2=3~Kg[/tex] the masses of the two carts, and [tex]v_1=6~m/s[/tex] and [tex]v_2[/tex] the velocities of the two carts before the collision, the initial momentum is [tex]p_1 = m_1 v_1 - m_2v_2[/tex] where the negative sign means the two carts are moving into opposite directions.
2) The two carts after collision are at rest, this means the total momentum after collision is [tex]p_2=0[/tex]
3) Since the momentum must be conserved, we can write [tex]p_1=p_2[/tex] and substituting [tex]m_1 v_1 - m_2v_2=0[/tex] from which we can get the velocity of the second cart: [tex]v_2= \frac{m_1}{m_2}v_1= \frac{2Kg}{3Kg} 6~m/s = 4~m/s [/tex]
