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Chapter 4: Problem 466

Two bodies \(\mathrm{P}\) and \(\mathrm{Q}\) have masses \(5 \mathrm{~kg}\) and $20 \mathrm{~kg}\( respectively. Each one is acted upon by a force of \)4 \mathrm{~N}$. If they acquire the same kinetic energy in times \(t_{\mathrm{P}}\) and \(\mathrm{t}_{\mathrm{Q}}\) then the ratio $\left(t_{q} / t_{p}\right)=\ldots \ldots$ \((\mathrm{A})(1 / 2)\) (B) 2 (C) 5 (D) 6

Short Answer

Expert verified
The short answer is: \(\frac{t_Q}{t_P} = \frac{2}{\sqrt{5}} = \boxed{(\mathrm{C})\ 5}\)

Step by step solution

01

Calculate Acceleration for Body P and Body Q

Recall the basic formula of force: \(F = ma\). The given constant force of 4N acts on both masses, we can find the acceleration for both by rearranging the formula: For Body P: \(a_P = \frac{F}{m_P} = \frac{4}{5}\) For Body Q: \(a_Q = \frac{F}{m_Q} = \frac{4}{20} = \frac{1}{5}\)
02

Derive Expressions for Final Velocity and Time

Using the equation of motion \(v = u + at\), and since both bodies start from rest (u = 0), we have: For Body P: \(v_P = a_P \cdot t_P = \frac{4}{5}t_P\) For Body Q: \(v_Q = a_Q \cdot t_Q = \frac{1}{5}t_Q\)
03

Calculate the Kinetic Energy of Both Bodies

Recall the equation of kinetic energy, \(KE = \frac{1}{2}mv^2\). We have: For Body P: \( KE_P = \frac{1}{2}(5)(v_P)^2 = \frac{1}{2}(5)\left(\frac{4}{5}t_P\right)^2\) For Body Q: \( KE_Q = \frac{1}{2}(20)(v_Q)^2 = \frac{1}{2}(20)\left(\frac{1}{5}t_Q\right)^2\)
04

Set the Kinetic Energy of Both Bodies Equal and Solve for the Ratio

Since the problem states that both bodies acquire the same kinetic energy, we can set KE_P equal to KE_Q and solve for the ratio \(t_Q / t_P\): \(\frac{1}{2}(5)\left(\frac{4}{5}t_P\right)^2 = \frac{1}{2}(20)\left(\frac{1}{5}t_Q\right)^2\) \(\frac{1}{2}\cdot\frac{16}{5}t_P^2 = \frac{1}{2}\cdot 4 t_Q^2\) \(\frac{16}{5}t_P^2 = 4 t_Q^2\) Now, divide both sides by 4: \(\frac{4}{5}t_P^2 = t_Q^2\) Now, take the square root of both sides to get: \(\frac{2}{\sqrt{5}}t_P = t_Q\) Finally, solve for the ratio \( t_Q / t_P = \frac{2}{\sqrt{5}}\): \(\frac{t_Q}{t_P} = \frac{2}{\sqrt{5}} = \boxed{(\mathrm{C})\ 5}\)

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Most popular questions from this chapter

Two bodies of masses \(\mathrm{m}\) and \(3 \mathrm{~m}\) have same momentum. their respective kinetic energies \(E_{1}\) and \(E_{2}\) are in the ratio..... (A) \(1: 3\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 6\)

A uniform chain of length \(\mathrm{L}\) and mass \(\mathrm{M}\) is lying on a smooth table and \((1 / 4)^{\text {th }}\) of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \(\mathrm{MgL} / 9\) (C) \(\mathrm{MgL} / 18\) (D) \(\mathrm{MgL} / 32\)

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: stopping distance \(=[\\{\) Kinetic energy \(\\} /\\{\) Stopping force \(\\}]\) Reason: Work done in stopping a body is equal to K.E. of the body. (A) 1 (B) 2 (C) 3 (D) 4

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A bullet of mass \(0.10 \mathrm{~kg}\) moving with a speed of $100 \mathrm{~m} / \mathrm{s}\( enters a wooden block and is stopped after a distance of \)0.20 \mathrm{~m}$. What is the average resistive force exerted by the block on the bullet ? (A) \(2.5 \times 10^{2} \mathrm{~N}\) (B) \(25 \mathrm{~N}\) (C) \(25 \times 10^{2} \mathrm{~N}\) (D) \(2.5 \times 10^{4} \mathrm{~N}\)
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