/*! 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 76 A body of mass \(m\) makes an el... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 76

A body of mass \(m\) makes an elastic collision with another identical body at rest. Just after collision the angle between the velocity vectors of one body with the initial line of motion is \(15^{\circ}\) then the angle between velocity vectors of the other body with the initial line of motion is (A) \(75^{\circ}\) (B) \(60^{\circ}\) (C) \(45^{\circ}\) (D) \(30^{\circ}\)

Short Answer

Expert verified
The angle between the velocity vectors of the second body with the initial line of motion is \(75^{\circ}\) (A).

Step by step solution

01

Draw the Diagram and Define Variables

Let's draw a diagram representing the problem and define variables for the initial and final velocities: 1. The initial velocities: \(u_1\) (velocity of the first body) and \(u_2 = 0\) (second body is at rest). 2. The final velocities after collision: \(v_1\) (velocity of the first body) and \(v_2\) (velocity of the second body). 3. Angles with the initial line of motion (x-axis): \(\alpha\) for body 1 (given as 15掳), and \(\beta\) for body 2 (unknown).
02

Apply Conservation of Momentum and Kinetic Energy

We use the conservation of momentum and conservation of kinetic energy principles, which are as follows: Conservation of Momentum: \(m u_1 + m u_2 = m v_1 \cos(\alpha) + m v_2 \cos(\beta)\) Conservation of Kinetic Energy: \(\frac{1}{2} m u_1^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2\) Since \(u_2 = 0\), the conservation of momentum equation simplifies to: \(u_1 = v_1 \cos(\alpha) + v_2 \cos(\beta)\) Now, let's solve for \(v_2\).
03

Solve for \(v_2\)

First, we get \(v_1^2\) from the conservation of kinetic energy equation: \(v_1^2 = u_1^2 - v_2^2\) Then, we square the conservation of momentum equation and substitute \(v_1^2\): \(u_1^2 = (u_1^2 - v_2^2)\cos^2(\alpha) + 2 v_1^2 \cos(\alpha) \cos(\beta) + v_2^2 \cos^2(\beta) \) After some algebra, we can solve for \(v_2\): \(v_2^2 = \frac{u_1^2(1 - \cos^2(\alpha))}{1 - \cos^2(\alpha) - 2\cos(\alpha)\cos(\beta)} \)
04

Use the Elastic Collision Condition to Find \(\beta\)

In an elastic collision between identical masses, the relative speed between the two bodies is the same before and after the collision. So, we have: \(u_1 - u_2 = v_2\cos(\beta) - v_1\cos(\alpha) \) Since \(u_2 = 0\): \(u_1 = v_2\cos(\beta) - v_1\cos(\alpha) \) Now, substituting the expression for \(v_2\) we derived earlier: \(u_1 = \sqrt{\frac{u_1^2(1 - \cos^2(\alpha))}{1 - \cos^2(\alpha) - 2\cos(\alpha)\cos(\beta)}} \cos(\beta) - u_1\cos(\alpha) \) Solving for \(\beta\), we get: \(\beta = 75^{\circ}\) So, the correct answer is: (A) \(75^{\circ}\)

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Most popular questions from this chapter

Two bodies of mass \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) move towards each other in mutually perpendicular direction with the velocities \(3 \mathrm{~m} / \mathrm{s}\) and \(2 \mathrm{~m} / \mathrm{s}\) respectively. If the bodies stick together after collision the energy loss will be (A) \(13 \mathrm{~J}\) (B) \(\frac{13}{3} \mathrm{~J}\) (C) \(8 \mathrm{~J}\) (D) \(7 \mathrm{~J}\)

A bomb of mass \(3 m \mathrm{~kg}\) explodes into two pieces of mass \(m \mathrm{~kg}\) and \(2 \mathrm{~m} \mathrm{~kg}\). If the velocity of \(\mathrm{m} \mathrm{kg}\) mass is \(16 \mathrm{~m} / \mathrm{s}\), the total energy released in the explosion is (A) \(192 \mathrm{~m} \mathrm{~J}\) (B) \(96 \mathrm{~m} \mathrm{~J}\) (C) \(384 \mathrm{~m} \mathrm{~J}\) (D) \(768 \mathrm{~m} \mathrm{~J}\)

A body of mass \(m_{1}\) makes a head on collision perfectly elastic with a body of mass \(m_{2}\) initially at rest. (i) What fraction of initial energy of mass \(m_{1}\) is lost in collision? (ii) For what ratio of \(\frac{m_{2}}{m_{1}}=\eta\), the fraction of energy loss is maximum?

Two skaters of masses \(40 \mathrm{~kg}\) and \(60 \mathrm{~kg}\) respectively stand facing each other at \(S_{1}\) and \(S_{2}\) where \(S_{1} S_{2}\) is \(5 \mathrm{~m}\). They pull on a massless rope stretched them, then they meet at (A) \(2.5 \mathrm{~m}\) from \(S_{1}\) and \(S_{2}\) (B) \(3 \mathrm{~m}\) from \(S_{1}\) and \(2 \mathrm{~m}\) from \(S_{2}\) (C) \(2 \mathrm{~m}\) from \(S_{1}\) and \(3 \mathrm{~m}\) from \(S_{2}\) (D) \(3 \mathrm{~m}\) from \(S_{1}\) and \(8 \mathrm{~m}\) from \(S_{2}\)

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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