KCET · Physics · Center of Mass Momentum and Collision
If the linear momentum of a body is increased by \(50 \%\), then the kinetic energy of that body increases by
- A \(100 \%\)
- B \(125 \%\)
- C \(225 \%\)
- D \(25 \%\)
Answer & Solution
Correct Answer
(B) \(125 \%\)
Step-by-step Solution
Detailed explanation
Kinetic energy of the body, \(K=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}\) Since, the mass remains constant
\(\therefore\) \(\mathrm{K} \propto \mathrm{p}^{2}\)
\(\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{p}_{2}^{2}}{\mathrm{p}_{1}^{2}}=\left[\frac{150}{100}\right]^{2}=\frac{9}{4} \)
\(100\left[\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}-1\right]=\left(\frac{9}{4}-1\right) \times 100=125 \%\)
\(\therefore\) \(\mathrm{K} \propto \mathrm{p}^{2}\)
\(\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{p}_{2}^{2}}{\mathrm{p}_{1}^{2}}=\left[\frac{150}{100}\right]^{2}=\frac{9}{4} \)
\(100\left[\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}-1\right]=\left(\frac{9}{4}-1\right) \times 100=125 \%\)
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