Question
A lighter object $A$ and a heavier object $B$ are initially at rest. Both are imparted the same linear momentum. Which will start with greater kinetic energy: $A$ or $B$ or both will start with the same energy?
[Ans:$A$]
[Ans:$A$]
✓
Answer
- Let $m_1$ and $m_2$ be the masses of light object $A$ and heavy object $B$ and $v_1$ and $v_2$ be their respective velocities.
- Since both are imparted with the same linear momentum,$m_1 v_1 = m_2 v_2$
- Kinetic energy of the lighter object A
$= K \cdot E _{\cdot A }=\frac{1}{2} m _2 v _1^2$
Kinetic energy of the heavier object B
$= K \cdot E _{ B }=\frac{1}{2} m _2 v _2^2$
$\begin{array}{ll}\therefore \quad \frac{K \cdot E_{\cdot A }}{ K \cdot E _{ B }}=\frac{\frac{1}{2} m _1 v _1^2}{\frac{1}{2} m _2 v _2^2}=\frac{\left( m _1 v _1\right)^2 / m _1}{\left( m _2 v _2\right)^2 / m _2} \\ \therefore \quad \frac{K \cdot E _{ A }}{ K \cdot E _{ B }}=\frac{ m _2}{ m _1} \quad \ldots \ldots\left[\because m _1 v _1= m _2 v _2\right]\end{array}$
- As $m_1 < m_2$, therefore $K.E._A > K.E._B$, i.e, the lighter body $A$ has more kinetic energy.
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