/*! 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 3 A particle of mass \(m\) moving ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

A particle of mass \(m\) moving eastward with a speed \(v\) collides with another particle of the same mass moving northward with the same speed \(v\). The two particles coalesce on collision. The new particle of mass \(2 \mathrm{~m}\) will move in the north-east direction with a velocity (A) \(\sqrt{2} v\) (B) \(\frac{v}{2}\) (C) \(\frac{v}{\sqrt{2}}\) (D) \(v\)

Short Answer

Expert verified
The new particle of mass \(2m\) will move in the north-east direction with a velocity of \(\boxed{\dfrac{v}{\sqrt{2}}}\), which corresponds to option \(C\).

Step by step solution

01

Identify known and unknown quantities

From the problem, we know that: 1. The mass of each particle: \(m\) 2. The speed of particle 1 (eastward): \(v\) 3. The speed of particle 2 (northward): \(v\) 4. The combined mass after collision: \(2m\) Our objective is to determine the north-eastward velocity of the new particle.
02

Write down the conservation of momentum equations

Since momentum is conserved, we can write the equations of conservation of momentum for the x-axis (eastward) and the y-axis (northward). For the x-axis, we have: \(m_1v_1 + m_2v_2 = M V_x \) For the y-axis, we have: \(m_3v_3 + m_4v_4 = M V_y \) Where \(m_1 = m_2 = m\), \(v_1 = v\), \(v_2 = 0\) (since the second particle is not moving eastward), \(m_3 = m_4 = m\), \(v_3 = 0\) (since the first particle is not moving northward), \(v_4 = v\), and \(M = 2m\).
03

Apply conservation of momentum

Now let's apply the conservation of momentum equations for both directions. For the x-axis: \(mv + m(0) = 2m V_x\) For the y-axis: \(m(0) + mv = 2m V_y\)
04

Solve for \(V_x\) and \(V_y\)

From the x-axis equation, we get: \(V_x = \frac{mv}{2m} = \frac{v}{2}\) From the y-axis equation, we get: \(V_y = \frac{mv}{2m} = \frac{v}{2}\)
05

Find the north-eastward velocity

To find the north-eastward velocity, we have to find the magnitude of the resultant vector. Using the Pythagorean theorem, we get: \(V = \sqrt{V_x^2 + V_y^2} = \sqrt{\left(\frac{v}{2}\right)^2 + \left(\frac{v}{2}\right)^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}}\) \(V = \frac{v}{\sqrt{2}}\) #Answer#: The new particle of mass \(2m\) will move in the north-east direction with a velocity of \(\boxed{\dfrac{v}{\sqrt{2}}}\), which corresponds to option \(C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Collision
When two particles collide, they come into contact with each other physically. In this scenario, both particles have the same mass and move towards each other from perpendicular directions鈥攅ast and north. After the collision, they combine to form one larger mass, indicating it is a completely inelastic collision.
Inelastic collisions occur when the colliding objects stick together after impact. Here:
  • Before collision: Two separate particles, each of mass \(m\), moving east and north with velocity \(v\).
  • After collision: A combined particle with mass \(2m\).
During such collisions, the momentum is conserved, meaning the total momentum before the collision leads to the momentum of the single mass after they stick together. Understanding this concept is key to solving momentum problems.
Resultant Velocity
After the collision, the combined mass has a new velocity. Because each original particle was moving perpendicular to the other, we can break down the velocities into components and use vector addition to find the resultant velocity.
The conservation of momentum equations help us determine these components:
  • Eastward (x-component): \(V_x = \frac{v}{2}\)
  • Northward (y-component): \(V_y = \frac{v}{2}\)
The resultant velocity is found by combining these components. The vector triangle method can be used, where the components form the legs of the right triangle, and the resultant velocity is the hypotenuse. This approach enables us to visualize and calculate how the combined particle moves in the north-east direction.
Pythagorean Theorem
The Pythagorean theorem is a crucial tool in calculating the magnitude of the resultant velocity. If you've ever worked with right triangles, this theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
For our problem:
  • The eastward and northward velocities, \(V_x\) and \(V_y\), act as legs of a right triangle.
  • The resultant velocity \(V\) is the hypotenuse.
Using the Pythagorean theorem, we have:\[V = \sqrt{V_x^2 + V_y^2} = \sqrt{\left(\frac{v}{2}\right)^2 + \left(\frac{v}{2}\right)^2} = \sqrt{\frac{v^2}{2}}\]This equation allows us to calculate the exact speed of the mass after impact, helping us understand how vector quantities combine in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 particle of mass \(m_{1}\) moving with velocity \(u_{1}\) strikes another particle of mass \(m_{2}\) moving with velocity \(u_{2}\). After collision the velocities of the particles are \(v_{1}\) and \(v_{2}\) respectively. Both the particles are moving on a frictionless horizontal surface. Column-I represents the nature of collision between the particles and Column-II represents the equations for physical quantities. (Symbols have their usual meanings) Column-I (A) Head on elastic collision (B) Head on inelastic collision (C) Oblique elastic collision (D) Oblique inelastic collision Column-II 1\. \(m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}\) \(=m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}\) 2\. \(e=\frac{\left|\vec{v}_{2}\right|+\left|\overrightarrow{v_{1}}\right|}{\left|\overrightarrow{u_{1}}\right|+\left|\overrightarrow{u_{2}}\right|}\) \(\begin{aligned} \text { 3. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\ &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\\ \text { 4. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\\ &>\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \end{aligned}\)

A machine gun fires a bullet of mass \(40 \mathrm{~g}\) with a velocity \(1200 \mathrm{~ms}^{-1}\). The man holding it can exert a maximum force of \(144 \mathrm{~N}\) on the gun. How many bullets can he fire per second at the most? (A) Two (B) Four (C) One (D) Three

A particle of mass \(m\) moving eastward with a speed \(v\) collides with another particle of same mass moving northward with same speed \(v .\) The two particles coalesce on collision. The new particle of mass \(2 \mathrm{~m}\) will move in the north-east direction with a velocity of (A) \(v \sqrt{2}\) (B) \(\frac{v}{\sqrt{2}}\) (C) \(\frac{v}{2}\) (D) \(v\)

If a metal wire is stretched a little beyond its elastic limit (or yield point) and released, it will (A) lose its elastic property completely. (B) not contract. (C) contract, but its final length will be greater than its initial length. (D) contract only up to its initial length at the elastic limit.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Waves Physics

Read Explanation

Radiation

Read Explanation

Space Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.