Three bodies A, B and C have equal kinetic en | vedclass

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(A) The kinetic energy $KE$ of a body is related to its linear momentum $P$ and mass $m$ by the formula $KE = \frac{P^2}{2m}$.Since the kinetic energi

Vedclass · Accepted Answer

$1: \sqrt{3}: 2$

Vedclass · Answer

(A) The kinetic energy $KE$ of a body is related to its linear momentum $P$ and mass $m$ by the formula $KE = \frac{P^2}{2m}$.Since the kinetic energies are equal, we have $P^2 = 2m(KE)$, which implies $P = \sqrt{2m(KE)}$.Since $2$ and $KE$ are constants, the linear momentum is directly proportional to the square root of the mass: $P \propto \sqrt{m}$.Given masses are $m_A = 400 \,g = 0.4 \,kg$, $m_B = 1.2 \,kg$, and $m_C = 1.6 \,kg$.The ratio of their momenta is $P_A : P_B : P_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$.Substituting the values: $P_A : P_B : P_C = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$.Multiplying by $\sqrt{10}$ to simplify: $\sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4$.Dividing by $2$, we get $1 : \sqrt{3} : 2$.

Three bodies $A, B$ and $C$ have equal kinetic energies and their masses are $400 \,g, 1.2 \,kg$ and $1.6 \,kg$ respectively. The ratio of their linear momenta is:

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