System energy before bullet impact: \text{System energy before bullet impact:} System energy before bullet impact:
E 1 = m V 1 2 2 + M g h 1 ; E_1 =\frac{mV_1^2}{2}+Mgh_1; E 1 = 2 m V 1 2 + M g h 1 ;
System energy after bullet impact: \text{System energy after bullet impact:} System energy after bullet impact:
E 2 = m V 2 2 2 + M g h 2 ; E_2 =\frac{mV_2^2}{2}+Mgh_2; E 2 = 2 m V 2 2 + M g h 2 ;
where: \text{where:} where:
m = 8.75 ∗ 1 0 − 3 k g − bullet mass m=8.75*10^{-3}kg-\text{bullet mass} m = 8.75 ∗ 1 0 − 3 k g − bullet mass
M = 0.705 k g ballistic pendulum mass M=0.705 \ kg\text{ballistic pendulum mass} M = 0.705 k g ballistic pendulum mass
V 1 , V 2 bullet velocity before and after impact V_1,V_2\text{ bullet velocity before and after impact} V 1 , V 2 bullet velocity before and after impact
h 1 , h 2 ballistic pendulum height above ground h 2 − h 1 = 0.2183 m h_1,h_2 \text{ ballistic pendulum height above ground }h_2-h_1=0.2183\ m h 1 , h 2 ballistic pendulum height above ground h 2 − h 1 = 0.2183 m
E 1 = E 2 E_1=E_2 E 1 = E 2
m V 1 2 2 + M g h 1 = m V 2 2 2 + M g h 2 \frac{mV_1^2}{2}+Mgh_1=\frac{mV_2^2}{2}+Mgh_2 2 m V 1 2 + M g h 1 = 2 m V 2 2 + M g h 2
m V 1 2 2 − m V 2 2 2 = M g h 2 − M g h 1 \frac{mV_1^2}{2}-\frac{mV_2^2}{2}=Mgh_2-Mgh_1 2 m V 1 2 − 2 m V 2 2 = M g h 2 − M g h 1
m V 2 2 2 = m V 1 2 2 − M g ( h 2 − h 1 ) \frac{mV_2^2}{2}=\frac{mV_1^2}{2}-Mg(h_2-h_1) 2 m V 2 2 = 2 m V 1 2 − M g ( h 2 − h 1 )
m V 2 2 2 = \frac{mV_2^2}{2}= 2 m V 2 2 =
V 2 = ( V 1 2 − 2 M g ( h 2 − h 1 ) m ) ≈ 396.56 V_2=\sqrt{(V_1^2-\frac{2Mg(h_2-h_1)}{m})}\approx396.56 V 2 = ( V 1 2 − m 2 M g ( h 2 − h 1 ) ) ≈ 396.56
Answer: 396.56m/s
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