Answer to Question #176256 in Classical Mechanics for Annie Porter

Question #176256

 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?


1
Expert's answer
2021-03-29T09:00:02-0400

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

E1=mV122+Mgh1;E_1 =\frac{mV_1^2}{2}+Mgh_1;

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

E2=mV222+Mgh2;E_2 =\frac{mV_2^2}{2}+Mgh_2;

where:\text{where:}

m=8.75103kgbullet massm=8.75*10^{-3}kg-\text{bullet mass}

M=0.705 kgballistic pendulum massM=0.705 \ kg\text{ballistic pendulum mass}

V1,V2 bullet velocity before and after impactV_1,V_2\text{ bullet velocity before and after impact}

h1,h2 ballistic pendulum height above ground h2h1=0.2183 mh_1,h_2 \text{ ballistic pendulum height above ground }h_2-h_1=0.2183\ m

E1=E2E_1=E_2

mV122+Mgh1=mV222+Mgh2\frac{mV_1^2}{2}+Mgh_1=\frac{mV_2^2}{2}+Mgh_2

mV122mV222=Mgh2Mgh1\frac{mV_1^2}{2}-\frac{mV_2^2}{2}=Mgh_2-Mgh_1

mV222=mV122Mg(h2h1)\frac{mV_2^2}{2}=\frac{mV_1^2}{2}-Mg(h_2-h_1)

mV222=\frac{mV_2^2}{2}=

V2=(V122Mg(h2h1)m)396.56V_2=\sqrt{(V_1^2-\frac{2Mg(h_2-h_1)}{m})}\approx396.56

Answer:396.56m/s




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