Answer to Question #176256 in Classical Mechanics for Annie Porter

Question

 A 8.75- g bullet from a 9-mm pistol has a velocity of 397.0 m/s. It strikes the 0.705- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 21.83 cm, what was the velocity of the bullet as it emerged from the block?

Answer 1

$\text{System energy before bullet impact:}$

$E_1 =\frac{mV_1^2}{2}+Mgh_1;$

$\text{System energy after bullet impact:}$

$E_2 =\frac{mV_2^2}{2}+Mgh_2;$

$\text{where:}$

$m=8.75*10^{-3}kg-\text{bullet mass}$

$M=0.705 \ kg\text{ballistic pendulum mass}$

$V_1,V_2\text{ bullet velocity before and after impact}$

$h_1,h_2 \text{ ballistic pendulum height above ground }h_2-h_1=0.2183\ m$

$E_1=E_2$

$\frac{mV_1^2}{2}+Mgh_1=\frac{mV_2^2}{2}+Mgh_2$

$\frac{mV_1^2}{2}-\frac{mV_2^2}{2}=Mgh_2-Mgh_1$

$\frac{mV_2^2}{2}=\frac{mV_1^2}{2}-Mg(h_2-h_1)$

$\frac{mV_2^2}{2}=$

$V_2=\sqrt{(V_1^2-\frac{2Mg(h_2-h_1)}{m})}\approx396.56$

Answer:396.56m/s