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Same force acts on two bodies of different masses \(2 \mathrm{~kg}\) and $4 \mathrm{~kg}$ initially at rest. the ratio of times required to acquire same final velocity is (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 1\) (D) \(4: 16\)

Short Answer

Expert verified
The ratio of times required for the two bodies to reach the same final velocity is \(1:2\), which corresponds to option (B).

Step by step solution

01

Understanding the given problem

We are given that the same force acts on two bodies with masses 2 kg and 4 kg which are initially at rest. We need to find the ratio of time taken for these two bodies to achieve the same final velocity.
02

Using force equation and kinematic equation

We will use the equation F = ma (Force = mass * acceleration) to find the acceleration of each body, where 'm' is the mass and 'a' is the acceleration. Additionally, we know that final velocity (v) can be determined using the kinematic equation: \(v = u + at\), where 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time taken.
03

Finding acceleration for each body using force equation

Since the same force acts on both bodies, we can denote the force as F. For the first body with mass 2 kg and acceleration \(a_1\): \(F = m_1a_1\) \(a_1 = \frac{F}{m_1}\) For the second body with mass 4 kg and acceleration \(a_2\): \(F = m_2a_2\) \(a_2 = \frac{F}{m_2}\)
04

Using the kinematic equation for final velocity of each body

We are given that both bodies reach the same final velocity 'v'. Since both bodies are initially at rest, their initial velocity (u) is 0. For the first body with acceleration \(a_1\) and time \(t_1\): \(v = 0 + a_1t_1\) \(t_1 = \frac{v}{a_1}\) For the second body with acceleration \(a_2\) and time \(t_2\): \(v = 0 + a_2t_2\) \(t_2 = \frac{v}{a_2}\)
05

Finding the ratio of times required to reach the same final velocity

Now, we need to find the ratio \(T = \frac{t_1}{t_2}\) Substituting the values of \(t_1\) and \(t_2\) using the information obtained in step 4: \(T = \frac{\frac{v}{a_1}}{\frac{v}{a_2}}\) \(T = \frac{a_2}{a_1}\) Substituting the values of \(a_1\) and \(a_2\) determined in step 3: \(T = \frac{\frac{F}{m_2}}{\frac{F}{m_1}}\) \(T = \frac{m_1}{m_2}\) Finally, substituting the values of \(m_1 = 2\ \mathrm{kg}\) and \(m_2 = 4\ \mathrm{kg}\): \(T = \frac{2\ \mathrm{kg}}{4\ \mathrm{kg}} = \frac{1}{2}\) So, the ratio of times required to reach the same final velocity is \(\boxed{1:2}\) which corresponds to option (B).

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