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Chapter 4: Problem 469

If the K.E. of a body is increased by \(44 \%\), its momentum will increase by....... (A) \(20 \%\) (B) \(22 \%\) (C) \(2 \%\) (D) \(120 \%\)

Short Answer

Expert verified
The momentum will increase by \(20\%\).

Step by step solution

01

Express kinetic energy in terms of momentum

We begin by writing the equation for the kinetic energy (K.E.) of a body. \[K.E. = \frac{1}{2}mv^2\] Next, we recall the equation for the momentum (p) of a body, which is given by: \[p = mv\] Now, we express the kinetic energy in terms of momentum. To do this, we square the momentum equation to obtain: \[p^2 = m^2v^2\] We then solve this equation for \(v^2\): \[v^2= \frac{p^2}{m^2}\] Now, substitute the expression for \(v^2\) in the kinetic energy equation: \[K.E. = \frac{1}{2m}\cdot\frac{p^2}{m}\]
02

Determine the percentage change in momentum

Let the initial kinetic energy be \(K.E._1\) and the final kinetic energy be \(K.E._2\). Similarly, the initial momentum is \(p_1\) and the final momentum is \(p_2\). We are given that the kinetic energy increases by 44%: \[K.E._2 = 1.44\cdot K.E._1\] Substitute the expressions for kinetic energy in terms of momentum, we get: \[\frac{1}{2m}\cdot\frac{p_2^2}{m} = 1.44 \cdot \frac{1}{2m}\cdot\frac{p_1^2}{m}\] Simplify: \[\frac{p_2^2}{p_1^2} = 1.44\] Now, we take the square root of both sides of the equation, and we get: \[\frac{p_2}{p_1} = 1.2\]
03

Calculate the percentage increase in momentum

Now that we found the ratio of the final to the initial momentum, we can calculate the percentage increase in momentum. The percentage increase is given by the formula: \[\% \text{ increase} = \frac{p_2 - p_1}{p_1} \times 100\%\] Using the ratio we found in the previous step: \[\% \text{ increase} = \frac{1.2p_1 - p_1}{p_1} \times 100\%\] Simplify the equation: \[\% \text{ increase} = 0.2 \times 100\% = 20\%\] So the correct answer is: (A) \(20\%\)

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