Percentage change in momentum for kinetic energy change – Rankers Physics

Kinetic Energy and Momentum: Practice Problem & Solution

If kinetic energy of a body is increased by \(300\%\) then percentage change in momentum will be: (2002)
\(100\%\)
\(150\%\)
\(265\%\)
\(73.2\%\)

Solution Explained:

To solve this problem, we apply the core principles of Kinetic Energy and Momentum. Understanding the underlying formula is key to arriving at the correct answer below:

Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). If \(KE_i\) is initial KE, then \(KE_f = KE_i + 300\%\ KE_i = 4KE_i\). So, \(p_f = \sqrt{2m(4KE_i)} = 2\sqrt{2mKE_i} = 2p_i\). Percentage change in momentum is \(\frac{p_f - p_i}{p_i} \times 100\% = \frac{2p_i - p_i}{p_i} \times 100\% = 100\%\).

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