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Solution).given Data;- Bulletmass (\mathrm{m})=12 \mathrm{~g}=.012 \mathrm{~kg} Block mass (M)=1.6 \mathrm{~kg}cord length (l)=2.2 \mathrm{~m}, \theta=60^{\circ}First We Need to calcolafed speed of (Blockt Bollet) just after impact Let it is (V).Applying Energy conservation.Total Energy jost aftersimpact of (B-l l e t+B lock) \approx Total energy of (Bullet + Block) at maximam Height\begin{array}{l}(K-E)_{I}+\underbrace{(P-E)^{\top}}_{(O)}=(K E)^{\top}+(P-E) \\\text { as } h=0(0) \\\text { as } V=0\end{array}(0)(0)us h=0 \quad as v=0\begin{aligned}\Rightarrow \frac{1}{2}(m+M) v^{2} & =(M+m) g h \\V & =\sqrt{2 g h}\end{aligned}where:- h (height) =l-l \cos \theta\begin{aligned}\therefore V= & \sqrt{2 g l(1-\cos 0)} \\= & \sqrt{2 \times 9.81 \times 2.2(1-6560)} \\& V=4.6456 \mathrm{~m} / \mathrm{s}\end{aligned}Now F.B.D of (Blockt mass)Cood tension:-T=(M+m) g+F_{c}centrifugal fore\begin{array}{c}T=(M+m) g+\frac{(M+m) v^{2}}{l} \quad(M+m) g F_{c} \\T=(1.6+.012) \times 9.81+\frac{(1.6+.012) \times(4.6456)^{2}}{2 \Omega} \\T=31.62 \mathrm{~N}\end{array} ...