Let
"m=0.012\\:\\rm kg"
"M=0.4\\:\\rm kg"
"v=70\\:\\rm m\/s"
The law of conservation of momentum gives
"mv=(m+M)u"So, the velocity of the block of wood when it strikes by a bullet
"u=\\frac{mv}{m+M}=\\frac{0.012\\times 70}{0.012+0.4}=2.04\\:\\rm m\/s"Since
"\\Delta E_{kinetic}=\\Delta E_{potential}"we get
"\\frac{(m+M)u^2}{2}=(m+M)gh"So
"h=\\frac{u^2}{2g}=\\frac{2.04^2}{2\\times9.8}=0.212\\:\\rm m"
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