/*! 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 147 A mass \(m\) moves with a veloci... [FREE SOLUTION] | 91Ó°ÊÓ

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A mass \(m\) moves with a velocity \(v\) and collides inelastically with another identical mass. After collision the first mass moves with velocity \(\frac{v}{\sqrt{3}}\) in a direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision. $$ \bullet \underset{m}{\longrightarrow} \quad \bullet v / \sqrt{3} $$ Before collision After collision (A) \(\sqrt{3} v\) (B) \(v\) (C) \(\frac{v}{\sqrt{3}}\) (D) \(\frac{2}{\sqrt{3}} v\)

Short Answer

Expert verified
The speed of the second mass after collision is (D) \(\frac{2}{\sqrt{3}}v\).

Step by step solution

01

Identify Known Quantities

We know: - Mass of both objects, \(m\) - Velocity of first mass before collision, \(v\) - Velocity of first mass after collision, \(v/\sqrt{3}\) - Direction of first mass after collision: Perpendicular to initial motion - We need to find the final speed of the second mass.
02

Conservation of Linear Momentum

Since no external forces are acting on the system, the linear momentum is conserved. We will address the conservation of momentum for x and y directions independently.
03

Momentum Conservation in the x-direction

Before collision, the momentum in the x-direction is \(mv\). After collision, the first mass is perpendicular to its initial motion, so its x-momentum is 0. Let the final x-component of the velocity of the second mass be \(v_x\). Then, we have: \[m(v) = m(0) + m(v_x) \Rightarrow v = v_x\]
04

Momentum Conservation in the y-direction

Before collision, the momentum in the y-direction is 0 (both masses were initially on the x-axis). After collision, the first mass is moving perpendicularly to the x-axis with velocity \(v/\sqrt{3}\). Let the final y-component of the velocity of the second mass be \(v_y\). Then, we have: \[m(0) + m(0) = m\left(\frac{v}{\sqrt{3}}\right) + m(v_y) \Rightarrow \frac{v}{\sqrt{3}} = v_y\]
05

Calculate the Final Velocity of the Second Mass

Now we can use the Pythagorean theorem to find the final velocity of the second mass. \(v_{2} = \sqrt{v_x^2 + v_y^2} = \sqrt{v^2 + \frac{v^2}{3}} = \sqrt{\frac{4v^2}{3}}\) \[\Rightarrow v_{2} = \frac{2}{\sqrt{3}}v\] So, the correct answer is (D) \(\frac{2}{\sqrt{3}}v\).

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Key Concepts

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

Momentum Conservation
In any collision, the principle of momentum conservation is fundamental. This principle states that the total momentum of a closed system remains constant if no external forces are acting. We can express it as:
  • Initial Total Momentum = Final Total Momentum
Let's break it down with an inelastic collision scenario where two masses collide and stick together or move together post-collision. Before the collision, each object has its own momentum, which is simply the product of its mass and velocity. After the collision, the sum of their momenta remains the same as it was before.

For the problem at hand, this means that we must consider both the x and y directions to fully account for how momentum is conserved. By applying this principle, we find separate equations for each direction. This ensures that we capture all the nuances of the momentum changes occurring during the inelastic collision.
Velocity Components
When analyzing collisions, especially in two dimensions, breaking down velocities into components is very useful. Each velocity has horizontal (x) and vertical (y) components. By doing this, analysis becomes simpler as it allows us to assess each dimension separately.

In the example from the exercise, the first mass initially moves horizontally and then switches to moving vertically after collision. We use the concept of velocity components to manage this movement change.
  • In the x-direction, the first mass's motion ends, so its entire x-component of velocity contributes to the second mass’s velocity.
  • In the y-direction, the first mass gains a velocity component perpendicular to the original direction.
By setting up equations for these directions, we can effectively keep track of how velocities change in each dimension. Velocity components serve to make calculations more orderly and intuitive.
Inelastic Collision Problems
Inelastic collision problems focus on the inability of objects involved to conserve kinetic energy, despite conserving momentum. In such collisions, some of the original kinetic energy is transformed into other forms of energy, such as heat or sound. This means that you cannot merely rely on kinetic energy calculations to find post-collision speeds.

Specifically, in an inelastic collision, like the one described, knowing the exact direction and speed of each object afterward isn't straightforward without calculating momentum in both axes.
  • The velocity of the first mass changes direction, highlighting the inelastic nature as kinetic energy is lost.
  • The problem is centered on calculating the precise speed of the second mass, relying on momentum conservation.
Steps such as identifying known quantities, addressing each velocity component, and using mathematical relations like the Pythagorean theorem, are key. Each of these tools help find the post-collision velocities when faced with inelastic collision challenges.

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Most popular questions from this chapter

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)

Statement-1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2: Principle of conservation of momentum holds true for all kinds of collisions. (A) Statement-1 is true, Statement- 2 is true; Statement- 2 is the correct explanation of Statement- 1 (B) Statement-1 is true, Statement- 2 is true; Statement- 2 is not the correct explanation of Statement-1 (C) Statement- 1 is false, Statement- 2 is true. (D) Statement- 1 is true, Statement- 2 is false.

A particle of mass \(2 \mathrm{~kg}\) starts moving in a straight line with an initial velocity of \(2 \mathrm{~m} / \mathrm{s}\) at a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). The rate of change of kinetic energy is (A) four times the velocity at any moment. (B) two times the displacement at any moment. (C) four times the rate of change of velocity at any moment. (D) constant throughout.

A particle of mass \(m\) is placed on top of a smooth hemispherical wedge of mass \(4 m\) at rest. The hemispherical wedge is kept on a frictionless horizontal surface. The particle is given a gentle push. The angular velocity of the particle relative to the wedge if wedge has velocity \(v\) when particle has fallen an angle \(\theta\) with respect to the wedge is (A) \(\frac{5 v}{R \cos \theta}\) (B) \(\frac{v}{R \cos \theta}\) (C) \(\frac{4 v}{R \cos \theta}\) (D) \(\frac{5 v}{R \sin \theta}\)

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\. \(e=\frac{\left|\vec{v}_{2}\right|+\left|\overrightarrow{v_{1}}\right|}{\left|\overrightarrow{u_{1}}\right|+\left|\overrightarrow{u_{2}}\right|}\) \(\begin{aligned} \text { 3. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\ &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\\ \text { 4. } & \frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} \\\ &>\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \end{aligned}\)

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