Final velocity=2m/s
kinetic energy lost = 96J
Explanation:
This question is an example for total inelastic collision.
For this type of collision, the equation is,
Step1: Finding the velocity
m1v1+m2v2 =(m1+m2) vf[tex]m_1v_1+m_2v_2 =(m_1+m_2) v_f[/tex]
4*8+12*0 = (4+12)v
32 = 16v
[tex]v= frac{32}{16}[/tex] = 2m/s
Step 2:Finding the kinetic energy
[tex]Kinetic energy = \frac{1}{2}\times mv^2[/tex]
[tex]\text{Kinetic Energy before collision} = \frac{1}{2} m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
Kinetic Energy before collision=[tex]\frac{1}{2}\times4\times8^2+\frac{1}{2}\times12\times0^2[/tex]
Kinetic Energy before collision= 128J
[tex]\text{Kinetic Energy after collision} = \frac{1}{2} m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]\text{ Kinetic Energy after collision}= \frac{1}{2}\times4\times2^2+\frac{1}{2}\times1\times*2^2[/tex]
Kinetic Energy after collison= 32J
Step 3: Finding the kinetic energy lost
Kinetic Energy lost = Kinetic Energy after collision – Kinetic Energy before collision
Kinetic energy lost=128 -32
Kinetic energy lost=96J
