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A body of mass \(20 \mathrm{~kg}\) is moving with a velocity of \(2 v\) and another body of mass \(10 \mathrm{~kg}\) is moving with velocity \(v\) along same direction. The velocity of their centre of mass is (A) \(\frac{5 v}{3}\) (B) \(\frac{2 v}{3}\) (C) \(\underline{y}\) (D) Zero

Short Answer

Expert verified
The velocity of their center of mass is \(\frac{5v}{3}\).

Step by step solution

01

Compute the product of mass and velocity for each body

First, we need to multiply each mass by its corresponding velocity: - For the first body, product = m1 * v1 = 20 kg * 2v = 40v - For the second body, product = m2 * v2 = 10 kg * v = 10v
02

Calculate the total mass of the system

Next, we need to add the masses of both bodies: Total mass = m1 + m2 = 20 kg + 10 kg = 30 kg
03

Compute the center of mass velocity

Now, we can calculate the center of mass velocity using the formula from the analysis section: V_cm = (m1 * v1 + m2 * v2) / (m1 + m2) V_cm = (40v + 10v) / 30 kg V_cm = 50v / 30 kg
04

Simplify the expression

We can simplify the expression for the center of mass velocity by dividing the numerator and denominator by 10: V_cm = 5v / 3 Comparing this result to the given options, the answer is: (A) \(\frac{5v}{3}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity
Velocity is a fundamental concept in physics that describes how fast something is moving and in which direction. It is different from speed, which only considers how fast something is moving without regard to direction. For instance, if a car is traveling north at 60 mph, its velocity is 60 mph northward.

The formula for velocity is:
  • Velocity = Displacement / Time
In the context of the exercise, two bodies with different velocities contribute to the velocity of their joint center of mass. One has a velocity of \(2v\) and the other \(v\), reflecting both speed and direction of movement.

Understanding velocity in the context of systems like this helps us predict the behavior of multiple moving objects as they interact.
Mass
Mass is a measure of the amount of matter in an object, and it doesn't change regardless of its location in the universe. It's a key property in determining how an object will respond to forces. In this problem, two different masses—20 kg and 10 kg—are considered.

In physics problems, mass is crucial for evaluating forces and movements. It influences how objects accelerate when subjected to forces, as defined by Newton's second law of motion:
  • Force = Mass \(\times\) Acceleration
Understanding mass helps us solve problems involving motion and forces in mechanical systems. In calculating the center of mass velocity, these specific masses factor into how the system's overall velocity is averaged.
Physics Problem Solving
Physics problem-solving involves applying principles and formulas systematically to understand how systems behave. Start by identifying what is given and what needs to be found. For the center of mass velocity problem, you're given the masses and velocities.

Break down the problem:
  • Calculate products of mass and velocity for each component.
  • Add the results together and divide by the total mass for the center of mass velocity.
Problem-solving in physics can be made easier by writing down known values and step-by-step working to ensure no part is missed. Remember to check your work with unit consistency and dimensional analysis.
Mechanical System Analysis
Mechanical system analysis involves understanding the motions and forces within a system. With multiple interacting parts, like the two masses in this exercise, it's essential to analyze how they work together.

For a system's center of mass, knowing how individual velocities and masses interact offers insight into the system's overall motion. This is crucial for designing machines or understanding natural systems like planetary orbits.

The center of mass calculation takes weighted averages of velocity based on mass. It's a simplification that offers a clear picture of the entire system's motion. By breaking down a system into its components, you can predict how changes in one part will affect the whole.

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Most popular questions from this chapter

A block of metal weighting \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is stuck by a jet releasing water at a rate of \(1 \mathrm{~kg} / \mathrm{sec}\) and at a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block will be (A) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(5.0 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(7.5 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(10 \mathrm{~m} / \mathrm{s}^{2}\)

Consider a block \(P\) of mass \(2 M\) placed on another block \(Q\) of mass \(4 M\) lying on a fixed rough horizontal surface of coefficient of friction \(\mu\) between this surface and the block \(Q\). Surfaces of \(P\) and \(Q\) interacting with each other are smooth. \(A\) mass \(M\) moving in the horizontal direction along a line passing through the centre of mass of block \(Q\) is normal to the face. Speed of mass \(M\) is \(v_{0}\) when it collides elastically with the block \(Q\) at a height \(s\) from the rough surface. Both the blocks have same length \(4 s\). Velocity \(v_{0}\) is enough to make the block \(P\) topple. When mass \(M\) collides with block \(Q\) with velocity \(v_{0}\) the block \(P\). (A) Does not move (B) Moves forward (C) Moves backward (D) Either (B) or (C)

A proton moving with velocity \(v\) collides elastically with a stationary \(\alpha\)-particle. The velocity of the proton after the collision is (A) \(-\frac{3 v}{5}\) (B) \(\frac{3 v}{5}\) (C) \(\frac{2 v}{5}\) (D) \(-\frac{2 v}{5}\)

A ball \(A\) of mass \(m\) moving with velocity \(u\), collides head on with another ball \(B\) of the same mass at rest. If the co-efficient of restitution between balls is \(e\), the ratio of the final and initial velocities of ball \(A\) will be (A) \(\frac{1+e}{1-e}\) (B) \(\frac{1+e}{2}\) (C) \(1-e\) (D) \(\frac{1-e}{2}\)

Position of two particles are given by \(x_{1}=2 t\) and \(x_{2}=2+3 t .\) The velocity of centre of mass at \(t=2 \mathrm{~s}\) is \(2 \mathrm{~m} / \mathrm{s}\). Velocity of centre of mass at \(t=4 \mathrm{~s}\) will be (A) \(2 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1 \mathrm{~m} / \mathrm{s}\) (D) Zero

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