Problem 333
Question
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) \(4: 16\)
Step-by-Step Solution
Verified Answer
The ratio of times required for the two bodies to reach the same final velocity is \(1:2\), which corresponds to option (B).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 \(a_2\):
\(F = m_2a_2\)
\(a_2 = \frac{F}{m_2}\)
4Step 4: 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 \(t_2\):
\(v = 0 + a_2t_2\)
\(t_2 = \frac{v}{a_2}\)
5Step 5: 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).
Key Concepts
Force and Mass RelationshipKinematics EquationsAcceleration CalculationTime Ratio in Motion
Force and Mass Relationship
The force and mass relationship is a fundamental part of Newton's Second Law of Motion, which can be expressed as the equation \( F = ma \). This formula tells us that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In simpler terms, if you apply the same amount of force to two objects of different masses, the one with less mass will accelerate more. This concept is crucial in understanding how different masses respond to the same force.
The exercise we're exploring helps illustrate this relationship. We have two bodies, one with a mass of 2 kg and the other with 4 kg. Both are subjected to the same force. According to \( F = ma \), the acceleration experienced by each body can be found by rearranging the formula to \( a = \frac{F}{m} \). Therefore, for a constant force, a heavier object (greater mass \( m \)) will have less acceleration, while a lighter object will experience more. This core understanding of how force and mass interact is central to solving the problem.
The exercise we're exploring helps illustrate this relationship. We have two bodies, one with a mass of 2 kg and the other with 4 kg. Both are subjected to the same force. According to \( F = ma \), the acceleration experienced by each body can be found by rearranging the formula to \( a = \frac{F}{m} \). Therefore, for a constant force, a heavier object (greater mass \( m \)) will have less acceleration, while a lighter object will experience more. This core understanding of how force and mass interact is central to solving the problem.
Kinematics Equations
Kinematics equations describe the motion of objects based on their initial velocity, acceleration, and the time involved. One common kinematic equation is \( v = u + at \), where:
In the given scenario, both bodies start from rest, which means the initial velocity \( u = 0 \). The equation simplifies to \( v = at \). This equation allows us to find the time taken to reach a certain final velocity, provided we know the acceleration, which in turn, depends on the force and mass relationship as explained earlier. By applying these fundamental kinematics concepts, we can determine how long each body takes to reach the desired speed.
- \(v\) is the final velocity,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time taken.
In the given scenario, both bodies start from rest, which means the initial velocity \( u = 0 \). The equation simplifies to \( v = at \). This equation allows us to find the time taken to reach a certain final velocity, provided we know the acceleration, which in turn, depends on the force and mass relationship as explained earlier. By applying these fundamental kinematics concepts, we can determine how long each body takes to reach the desired speed.
Acceleration Calculation
Acceleration is the rate at which an object changes its velocity. It is a key concept when solving problems involving forces and motion. In the context of our exercise, it's essential to calculate the acceleration for each body to find the time it takes to acquire the same final velocity.
Given the force \( F \) is constant for both bodies, we use the rearranged version of Newton's Second Law, \( a = \frac{F}{m} \), to determine the acceleration:
Given the force \( F \) is constant for both bodies, we use the rearranged version of Newton's Second Law, \( a = \frac{F}{m} \), to determine the acceleration:
- For the 2 kg body, the acceleration \( a_1 = \frac{F}{2} \).
- For the 4 kg body, the acceleration \( a_2 = \frac{F}{4} \).
Time Ratio in Motion
The time ratio in motion is a way to compare how long it takes different objects to reach a particular state under certain conditions. In our problem, we want to find out how the times compare for two masses achieving the same final velocity when acted upon by the same force.
First, identify the expression for time in terms of acceleration and velocity from the kinematic equation \( v = at \):
\( t = \frac{v}{a} \). By substituting the expressions for acceleration (\( a_1\) and \( a_2\)) calculated previously:
First, identify the expression for time in terms of acceleration and velocity from the kinematic equation \( v = at \):
\( t = \frac{v}{a} \). By substituting the expressions for acceleration (\( a_1\) and \( a_2\)) calculated previously:
- The time for the 2 kg body, \( t_1 = \frac{v}{a_1} = \frac{v}{F/2} \).
- The time for the 4 kg body, \( t_2 = \frac{v}{a_2} = \frac{v}{F/4} \).
Other exercises in this chapter
Problem 330
A Particle moves in the X-Y plane under the influence of a force such that its linear momentum is \(\mathrm{P}^{-}(\mathrm{t})=\mathrm{A}[\mathrm{i} \cos (\math
View solution Problem 331
Force of \(5 \mathrm{~N}\) acts on a body of weight \(9.8 \mathrm{~N}\). what is the acceleration produced in \(\mathrm{ms}^{-2}\). (A) \(49.00\) (B) \(5.00\) (
View solution Problem 335
10,000 small balls, each weighing \(1 \mathrm{~g}\) strike one square \(\mathrm{cm}\) of area per second with a velocity \(100 \mathrm{~ms}^{-1}\) in a normal d
View solution Problem 338
A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at the rate of \(20 \mathrm{~ms}^{-1}\). If the catching process be completed in \(0.1 \mathrm
View solution