Question:
A rubber band catapult has initial length \(2 \, \text{cm}\) and cross-sectional area \(5 \, \text{mm}^2\). It is stretched to \(2 \, \text{cm}\) and then released to project a stone of mass \(20 \, \text{g}\). The velocity of projected stone is \((Y=5\times 10^8\,\text{N m}^{-2})\):
A rubber band catapult has initial length \(2 \, \text{cm}\) and cross-sectional area \(5 \, \text{mm}^2\). It is stretched to \(2 \, \text{cm}\) and then released to project a stone of mass \(20 \, \text{g}\). The velocity of projected stone is \((Y=5\times 10^8\,\text{N m}^{-2})\):
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For a stretched wire or rubber band, elastic potential energy is given by
\[
U=\frac{1}{2}\frac{YA(\Delta L)^2}{L}
\]
and this energy converts into kinetic energy when released.
Updated On: Jun 26, 2026
- \(20\,\text{m s}^{-1}\)
- \(50\,\text{m s}^{-1}\)
- \(100\,\text{m s}^{-1}\)
- \(250\,\text{m s}^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for elastic potential energy.
Elastic potential energy stored in a stretched rubber band is \[ U=\frac{1}{2}\frac{YA(\Delta L)^2}{L} \] Here, \[ Y=5\times 10^8\,\text{N m}^{-2} \] \[ A=5\,\text{mm}^2=5\times 10^{-6}\,\text{m}^2 \] \[ L=2\,\text{cm}=2\times 10^{-2}\,\text{m} \] \[ \Delta L=2\,\text{cm}=2\times 10^{-2}\,\text{m} \]
Step 2: Substitute the values.
\[ U=\frac{1}{2}\times \frac{(5\times 10^8)(5\times 10^{-6})(2\times 10^{-2})^2}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times \frac{(2500)(4\times 10^{-4})}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times \frac{1}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times 50 \] \[ U=25\,\text{J} \]
Step 3: Equate elastic potential energy to kinetic energy.
When the rubber band is released, its elastic potential energy is converted into kinetic energy of the stone.
So, \[ \frac{1}{2}mv^2=25 \] Mass of stone is \[ m=20\,\text{g}=20\times 10^{-3}\,\text{kg}=0.02\,\text{kg} \] Thus, \[ \frac{1}{2}\times 0.02 \times v^2=25 \] \[ 0.01v^2=25 \] \[ v^2=2500 \] \[ v=50\,\text{m s}^{-1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{50\,\text{m s}^{-1}} \]
Elastic potential energy stored in a stretched rubber band is \[ U=\frac{1}{2}\frac{YA(\Delta L)^2}{L} \] Here, \[ Y=5\times 10^8\,\text{N m}^{-2} \] \[ A=5\,\text{mm}^2=5\times 10^{-6}\,\text{m}^2 \] \[ L=2\,\text{cm}=2\times 10^{-2}\,\text{m} \] \[ \Delta L=2\,\text{cm}=2\times 10^{-2}\,\text{m} \]
Step 2: Substitute the values.
\[ U=\frac{1}{2}\times \frac{(5\times 10^8)(5\times 10^{-6})(2\times 10^{-2})^2}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times \frac{(2500)(4\times 10^{-4})}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times \frac{1}{2\times 10^{-2}} \] \[ U=\frac{1}{2}\times 50 \] \[ U=25\,\text{J} \]
Step 3: Equate elastic potential energy to kinetic energy.
When the rubber band is released, its elastic potential energy is converted into kinetic energy of the stone.
So, \[ \frac{1}{2}mv^2=25 \] Mass of stone is \[ m=20\,\text{g}=20\times 10^{-3}\,\text{kg}=0.02\,\text{kg} \] Thus, \[ \frac{1}{2}\times 0.02 \times v^2=25 \] \[ 0.01v^2=25 \] \[ v^2=2500 \] \[ v=50\,\text{m s}^{-1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{50\,\text{m s}^{-1}} \]
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