Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The object $A$ has half the kinetic energy as that of the object $B$ . The object $B$ has half the mass as that of the object $A$ . The object $A$ speeds up by $1\text{ ms}^{-1}$ and then has the same kinetic energy as that of the object $B$ . The initial speed of the object $A$ is: (Take $\sqrt{2} \cong 1.4$ )

$0.5\text{ ms}^{-1}$

$1\text{ ms}^{-1}$

$2.5\text{ ms}^{-1}$

$4.8\text{ ms}^{-1}$

Step-by-Step Solution

Identify Formula: Kinetic energy is given by $K = \frac{1}{2}mv^2$ [Class 11 Physics, Ch 6, Sec 5.4, Eq 5.5].
Set up Initial Conditions:

Mass relation: $m_B = \frac{m_A}{2}$ or $m_A = 2m_B$ .
Energy relation: $K_A = \frac{1}{2}K_B$ . $\frac{1}{2}m_A v_A^2 = \frac{1}{2} \left( \frac{1}{2}m_B v_B^2 \right)$ Substitute $m_A = 2m_B$ : $\frac{1}{2}(2m_B)v_A^2 = \frac{1}{4}m_B v_B^2 \implies 2v_A^2 = \frac{1}{2}v_B^2 \implies v_B^2 = 4v_A^2 \implies v_B = 2v_A$

Analyze Final Condition:

Object A speeds up: $v'_A = v_A + 1$ .
New Kinetic Energy equals $K_B$ : $K'_A = K_B$ . $\frac{1}{2}m_A (v_A + 1)^2 = \frac{1}{2}m_B v_B^2$ Substitute $m_A = 2m_B$ : $2(v_A + 1)^2 = v_B^2$

Solve System: Substitute $v_B = 2v_A$ into the second equation: $2(v_A + 1)^2 = (2v_A)^2 = 4v_A^2$ $(v_A + 1)^2 = 2v_A^2$ $v_A + 1 = \sqrt{2}v_A$ $1 = (\sqrt{2} - 1)v_A$ $v_A = \frac{1}{1.4 - 1} = \frac{1}{0.4} = 2.5\text{ ms}^{-1}$

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