A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1 − 1 into a block of mass 0.95kg. Given that the lead block can move freely, the final kinet...

A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1 ${}^{-1}$ into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?

Answer Details

Given 
m1 ${}_1$ = 0.05kg, u1 ${}_1$ = 200ms−1 ${}^{-1}$, m2 ${}_2$ = 0.95kg

K.E = 12 $\frac{1}{2}$mr ${}_r$v2 ${}^2$

m1 ${}_1$u1 ${}_1$ = v(m1 ${}_1$ + m2 ${}_2$) [law of conversation of momentum]

v = 0.05×2000.05+95 $\frac{0.05\times 200}{0.05+95}$ = 10ms−1 ${}^{-1}$

K.E = 12 $\frac{1}{2}$(1)102 ${}^2$ = 50J