MHT CET · Physics · Work Power Energy
Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio
of the magnitudes of their momenta is
- A \(4: 1\)
- B \(\sqrt{2}: 1\)
- C \(1: 12\)
- D \(1: 16\)
Answer & Solution
Correct Answer
(C) \(1: 12\)
Step-by-step Solution
Detailed explanation
Kinetic energy \(\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}\)
Two masses are given as- \(\mathrm{m}_{1}=1 \mathrm{gm}\) and \(\mathrm{m}_{2}=4 \mathrm{gm}\)
But \(\mathrm{K}_{1}=\mathrm{K}_{2}\)
\(\therefore \frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}=\frac{\mathrm{P}_{2}^{2}}{2 \mathrm{~m}_{2}}\)
Or \(\frac{\mathrm{P}_{1}^{2}}{2(1)}=\frac{\mathrm{P}_{2}^{2}}{2(4)}\)
\(\Longrightarrow \mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
Two masses are given as- \(\mathrm{m}_{1}=1 \mathrm{gm}\) and \(\mathrm{m}_{2}=4 \mathrm{gm}\)
But \(\mathrm{K}_{1}=\mathrm{K}_{2}\)
\(\therefore \frac{\mathrm{p}_{1}^{2}}{2 \mathrm{~m}_{1}}=\frac{\mathrm{P}_{2}^{2}}{2 \mathrm{~m}_{2}}\)
Or \(\frac{\mathrm{P}_{1}^{2}}{2(1)}=\frac{\mathrm{P}_{2}^{2}}{2(4)}\)
\(\Longrightarrow \mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
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