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)

CONSERVATION of LINEAR MOMENTUM
(m1v1i)i+(m2v2i)=(m1vif)+(m2v2f)
since they the collision is inelastic
v1f=v2f=vf
making the equation
(m1v1i)+(m2v2i)=(m1+m2)vf
(1.7kg)(1.5m/s)+(3.5kg)(-7.5m/s)=(1.7k… + 3.5kg)vf
-23.7kgm/s=5.2kg(vf)
vf=(-23.7kgm/s)/(5.2kg)
***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
and
KE=(1/2)m2v2f-m2v2i----->for KE lost in particle 2
Hope this helps you!
^^

Do your own homework!

do it by law of conservation of mass