Question

Bucket?

Answer 1

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$m_1v_1+m_2v_2 =(m_1+m_2) v_f$

4*8+12*0 = (4+12)v

32 = 16v

$v= frac{32}{16}$ = 2m/s

Step 2:Finding the kinetic energy

$Kinetic energy = \frac{1}{2}\times mv^2$

$\text{Kinetic Energy before collision} = \frac{1}{2} m_1v_1^2+\frac{1}{2}m_2v_2^2$

Kinetic Energy before collision=$\frac{1}{2}\times4\times8^2+\frac{1}{2}\times12\times0^2$

Kinetic Energy before collision= 128J

$\text{Kinetic Energy after collision} = \frac{1}{2} m_1v_1^2+\frac{1}{2}m_2v_2^2$

$\text{ Kinetic Energy after collision}= \frac{1}{2}\times4\times2^2+\frac{1}{2}\times1\times*2^2$

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