A rubber band catapult has initial length $2 \mathrm{~cm}$ and cross-sectional area $5 \mathrm{~mm}^2$. It…
A rubber band catapult has initial length $2 \mathrm{~cm}$ and cross-sectional area $5 \mathrm{~mm}^2$. It is stretched to $2 \mathrm{~cm}$ and then released to project a stone of mass of $20 \mathrm{~g}$. The velocity of projected stone is
(Young's modulus of rubber $=5 \times 10^8 \mathrm{Nm}^{-2}$ )
$20 \mathrm{~ms}^{-1}$
$50 \mathrm{~ms}^{-1}$
$100 \mathrm{~ms}^{-1}$
$250 \mathrm{~ms}^{-1}$
Solution
According to work-energy theorem, $\mathrm{K} \mathrm{E}$ of stone $=$ Elastic potential energy of rubber band.
Now, for rubber band;
Young's modulus $=5 \times 10^8 \mathrm{~N} / \mathrm{m}^2$
Length, $L=2 \times 10^{-2} \mathrm{~m}$
Change of length, $\Delta L=2 \times 10^{-2} \mathrm{~m}$
Area of cross-section, $A=5 \times 10^{-6} \mathrm{~m}^2$
Elastic potential energy of rubber band
$\begin{aligned} & =\frac{1}{2} \times Y \times(\text { strain })^2 \times \text { volume } \\ & =\frac{1}{2} \times Y \times\left(\frac{\Delta L}{L}\right)^2 \times A \times L \\ & =\frac{1}{2} \times 5 \times 10^8 \times\left(\frac{2 \times 10^{-2}}{2 \times 10^{-2}}\right)^2 \times 5 \times 10^{-6} \times 2 \times 10^{-2} \\ & =25 \mathrm{~J}\end{aligned}$
This energy is given to stone (mass $=20 \mathrm{~g}$ );
$\Rightarrow \mathrm{KE}$ (stone) = E.P.E (Rubber band)
$\Rightarrow \quad \frac{1}{2} \times m v^2=25$
$\Rightarrow \frac{1}{2} \times 20 \times 10^{-3} \times v^2=25$
$\begin{array}{ll}\Rightarrow & v^2=\frac{25 \times 2}{20 \times 10^{-3}} \Rightarrow v^2=25 \times 10^2 \\ \Rightarrow & v=5 \times 10=50 \mathrm{~m} / \mathrm{s}\end{array}$