KCET · Physics · Center of Mass Momentum and Collision
Two bodies with kinetic energies in the ratio of \(3:1\) are moving with equal linear momentum. The ratio of their masses is
- A \(1:4\)
- B \(1:3\)
- C \(1:2\)
- D \(1:1\)
Answer & Solution
Correct Answer
(B) \(1:3\)
Step-by-step Solution
Detailed explanation
The kinetic energy \(K\) of a body is related to its linear momentum \(p\) and mass \(m\) by the relation:
\(K = \dfrac{p^2}{2m}\)
Since the linear momenta of the two bodies are equal (\(p_1 = p_2 = p\)), the kinetic energy is inversely proportional to the mass:
\(K \propto \dfrac{1}{m}\)
Therefore, the ratio of their masses is:
\(\dfrac{m_1}{m_2} = \dfrac{K_2}{K_1}\)
Given that the ratio of their kinetic energies is \(\dfrac{K_1}{K_2} = \dfrac{3}{1}\), we get:
\(\dfrac{m_1}{m_2} = \dfrac{1}{3}\)
The ratio of their masses is \(1:3\).
Answer: \(1:3\)
\(K = \dfrac{p^2}{2m}\)
Since the linear momenta of the two bodies are equal (\(p_1 = p_2 = p\)), the kinetic energy is inversely proportional to the mass:
\(K \propto \dfrac{1}{m}\)
Therefore, the ratio of their masses is:
\(\dfrac{m_1}{m_2} = \dfrac{K_2}{K_1}\)
Given that the ratio of their kinetic energies is \(\dfrac{K_1}{K_2} = \dfrac{3}{1}\), we get:
\(\dfrac{m_1}{m_2} = \dfrac{1}{3}\)
The ratio of their masses is \(1:3\).
Answer: \(1:3\)
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