KCET · Physics · Center of Mass Momentum and Collision
Two bodies of masses \(m_{1}\) and \(m_{2}\) are acted upon by a constant force \(\mathrm{F}\) for a time t. They start from rest and acquire kinetic energies, \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\) respectively. Then \(\frac{E_{1}}{E_{2}}\) is
- A \(\frac{\sqrt{m_{1} m_{2}}}{m_{1}+m_{2}}\)
- B \(\frac{m_{1}}{m_{2}}\)
- C \(\frac{m_{2}}{m_{1}}\)
- D 1
Answer & Solution
Correct Answer
(C) \(\frac{m_{2}}{m_{1}}\)
Step-by-step Solution
Detailed explanation
Momentum acquired by the bodies
\(\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{Ft}\)
Now, their kinetic energies
\(\text {and } \mathrm{E}_{1}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}} \)
\(\text {and } \mathrm{E}_{2}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}} \)
\(\therefore \mathrm{E}_{1}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\)
\(\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{Ft}\)
Now, their kinetic energies
\(\text {and } \mathrm{E}_{1}=\frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}} \)
\(\text {and } \mathrm{E}_{2}=\frac{\mathrm{p}_{2}^{2}}{2 \mathrm{~m}_{2}} \)
\(\therefore \mathrm{E}_{1}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\)
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