The object $A$ has half the kinetic energy as that of the object $B$. The object $B$ has half the mass as that

Question

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

Accepted Answer

1. **Identify Formula:** Kinetic energy is given by $K = \frac{1}{2}mv^2$ [Class 11 Physics, Ch 6, Sec 5.4, Eq 5.5].
2. **Set up Initial Conditions:**
   - Mass relation: $m_B = \frac{m_A}{2}$ or $m_A = 2m_B$.
   - Energy relation: $K_A = \frac{1}{2}K_B$.
   $$ \frac{1}{2}m_A v_A^2 = \frac{1}{2} \left( \frac{1}{2}m_B v_B^2 ight) $$
   Substitute $m_A = 2m_B$:
   $$ \frac{1}{2}(2m_B)v_A^2 = \frac{1}{4}m_B v_B^2 \implies 2v_A^2 = \frac{1}{2}v_B^2 \implies v_B^2 = 4v_A^2 \implies v_B = 2v_A $$
3. **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 $m_A = 2m_B$:
   $$ 2(v_A + 1)^2 = v_B^2 $$
4. **Solve System:**
   Substitute $v_B = 2v_A$ into the second equation:
   $$ 2(v_A + 1)^2 = (2v_A)^2 = 4v_A^2 $$
   $$ (v_A + 1)^2 = 2v_A^2 $$
   $$ v_A + 1 = \sqrt{2}v_A $$
   $$ 1 = (\sqrt{2} - 1)v_A $$
   $$ v_A = \frac{1}{1.4 - 1} = \frac{1}{0.4} = 2.5	ext{ ms}^{-1} $$

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