/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A thin uniform bar lies on a fri... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is \(0.16 \mathrm{~kg}\) and length is \(1.7 \mathrm{~m}\). Two particles each of mass \(0.08 \mathrm{~kg}\) are moving on the same surface and towards the bar in the direction perpendicular to the bar, orre with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) and other with velocity 6\. \(\mathrm{m} / \mathrm{s}\). If collision between particles and bar is completely inelastic, both particles strike with the bar simultaneously. The velocity of centre of mass after collision is : (a) \(2 \mathrm{~m} / \mathrm{s}\) (b) \(4 \mathrm{~m} / \mathrm{s}\) (c) \(10 \mathrm{~m} / \mathrm{s}\) (d) \(16 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
Velocity of center of mass after collision is 4 m/s.

Step by step solution

01

Understand the Problem

We are given a uniform bar on a frictionless surface with two particles moving towards the bar. The bar's mass is \(0.16 \mathrm{~kg}\), with each particle having a mass of \(0.08 \mathrm{~kg}\). The problem specifies that both particles collide inelastically with the bar, meaning they stick to the bar post-collision. We need to find the velocity of the combined system's center of mass after the collision.
02

Identify the Conservation Principle

Since the collision is inelastic and occurs on a frictionless surface, we apply the law of conservation of linear momentum. The total linear momentum before the collision must equal the total linear momentum after the collision.
03

Calculate Initial Momentum

The total initial momentum \( p_{initial} \) is the sum of the individual momenta of the particles, because the bar is initially at rest.\( p_{initial} = m_1 \cdot v_1 + m_2 \cdot v_2 = 0.08 \cdot 10 + 0.08 \cdot 6 = 0.8 + 0.48 = 1.28 \mathrm{~kg~m/s}\).
04

Calculate Final Momentum

After the collision, the total mass of the system is the mass of the bar plus the masses of the two particles. This is \(0.16 + 0.08 + 0.08 = 0.32 \mathrm{~kg}\). Let \( v \) be the velocity of the center of mass after the collision. According to conservation of momentum,\( p_{final} = (0.32~ \mathrm{kg}) \cdot v\).
05

Set Initial Momentum Equal to Final Momentum

Set \( p_{initial} = p_{final} \) as per the conservation of momentum principle: \( 1.28 = 0.32 \cdot v \).
06

Solve for Final Velocity

Solve the equation \( 0.32v = 1.28 \) for \( v \). Divide both sides by \( 0.32 \) to get \( v = \frac{1.28}{0.32} = 4 \mathrm{~m/s}\). Thus, the final velocity of the center of mass after the collision is \(4 \mathrm{~m/s}\).

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is a fundamental concept in physics, especially useful in problems involving collisions. It states that the total momentum of a closed system remains constant if no external forces are acting on it. In simpler terms, what this means is that the momentum before an event, such as a collision, must equal the momentum after the event.

In this exercise, two particles are colliding with a bar on a frictionless surface. Frictionless means there are no external forces impacting the movement of the objects. Therefore, we can apply the conservation of momentum here.
  • First, compute the momentum of each moving object by multiplying its mass by its velocity.
  • Combine the individual momenta to find the total initial momentum of the system.
After the collision, since the objects stick together, you treat them as a single system. The total mass is the sum of individual masses, and using the conservation of momentum, the total initial momentum is set equal to the total final momentum. Solving this allows us to find the velocity of the center of mass after the collision.
Center of Mass
The center of mass of a system is a specific point where the system's mass can be considered to be concentrated. In collision problems, especially those involving multiple objects, the center of mass helps simplify the problem by focusing on a single point that moves according to the laws of physics.

In this problem, calculating the center of mass was necessary after the collision to determine its velocity.
  • Start by determining the combined mass: add up the mass of the bar and the masses of the two particles.
  • The velocity of the center of mass is derived using the conservation of momentum, which simplifies tracking how the momentum is transferred through the collision.
The center of mass moves at a constant velocity along the direction of total momentum, which is especially straightforward in cases of inelastic collisions, where objects stick together post-collision. Identifying the velocity of this point helps in describing the motion of the entire system after the impact.
Physics Problem Solving
Physics problem solving involves understanding concepts, applying laws, and performing calculations systematically. In the given exercise, several key steps were taken to find the solution.

The steps include:
  • Understanding the problem statement: Identify given values, like the mass and velocities of the objects involved.
  • Selecting the right physical principles: In this case, the conservation of momentum was crucial due to the nature of the collision and the frictionless environment.
  • Performing accurate calculations: Calculate initial momentum, then relate it to final momentum.
Being thorough in conceptual understanding allows you to choose suitable formulas and solve efficiently. Practicing these kinds of problems improves your skills in strategic problem solving and logical reasoning. This approach of decomposing the problem into manageable steps makes even complicated scenarios more approachable.

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Most popular questions from this chapter

4\. A non-uniform thin rod of length \(L\) is placed along \(X\) -axis as such its one of end is at the origin. The linear mass density of rod is \(\lambda=\lambda_{0} x .\) The distance of centre of mass of rod from the origin is : (a) \(\frac{L}{2}\) (b) \(\frac{2 L}{2}\) (c) \(\frac{L}{4}\) (d) \(\frac{L}{5}\)

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A block of mass \(M\) lying on a smooth horizontal surface is rigidly attached to a light horizontal spring of force constant \(k\). The other end of the spring is rigidly connected to a fixed wall. A stationary gun fires bullets of mass \(m\) each in horizontal direction with speed \(v_{0}\) one after other. The bullets hit the block and get embedded in it.The first bullet hits the block at \(t=0\). The second bullet hits at \(t=2 \pi \sqrt{\frac{M+m}{k}}\), the third bullet hits at \(t=2 \pi \sqrt{\frac{M+m}{k}}+2 \pi \sqrt{\frac{M+2 m}{k}}\) and so on. The maximum compression in the spring after the \(n\) th bullet hits is: (a) \(\frac{n m v_{0} \sqrt{k}}{(M+n m)^{3 / 2}}\) (b) \(\frac{(M+n m)^{3 / 2}}{n m v_{0} \sqrt{k}}\) (c) \(\frac{\sqrt{n m v_{0} k}}{(M+n m)^{3 / 2}}\) (d) \(\frac{n m v_{0}}{\sqrt{k(M+n m)}}\)

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