Question

A bullet of mass m moving with velocity v strikes a block of mass m at rest and gets embedded the loss of kinetic energy

Answer 1

Answer:

The loss of kinetic energy is $[\frac{m M \, u^{2} }{2 ( m+M)} ]$

Explanation:

Given;-

      Mass of bullet = m

      Mass of block = M

      Relative velocity = u

Now, we know that for a perfectly inelastic collision, the loss in the kinetic energy is given by;-

     Δ K.E. =  $\frac{1}{2} \frac{m1m2}{m1+m2} u^{2} relative$

So, putting the given vales in the formula, we get;-

     Δ K.E. = $\frac{1}{2}\frac{mM\, u^{2}}{m + M}$

     Δ K.E. =  $[\frac{mM\, u^{2} }{2(m + M)} ]$

Hope it helps! ;-))