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College Physics (Urone)
Found in: Page 291
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

Confirm that the results of the example 8.7 and do conserve momentum in both the x - and y -directions

In both the cases momentum is conserved.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of Linear Momentum

The product of an object's mass and velocity is called linear momentum, translational momentum, or simply momentum. It's a two-dimensional vector quantity with a magnitude and a direction.

The mass of the first one is\({m_1} = 0.250\;kg\).

The mass of the second one is\({m_2} = 0.400\;kg\).

The initial velocity of the first one before collision is\({\theta _2} = 45^\circ \).

Step 2: Linear momentum along X-axis

We can write along X-axis,

Linear momentum before collision is,

\(\begin{array}{c}mv = 0.250 \times 2\\ = 0.500\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Linear momentum after collision is,

\(\begin{array}{l}{m_1}{{v'}_1}\cos {\theta _1} + {m_2}{{v'}_2}\cos {\theta _2}\\ = \left[ {0.250 \times 1.50 \times \cos 55^\circ + 0.400 \times 896 \times \cos 312^\circ } \right]\;{\rm{kg}} \cdot {\rm{m/s}}\\ = 0.500\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Both are the same

Step 3: Linear momentum along Y-axis

We can write along Y-axis,

Linear momentum before collision is zero as the particles are moving along the X-axis.

Linear momentum after collision is,

\(\begin{array}{l}{m_1}{{v'}_1}\sin {\theta _1} + {m_2}{{v'}_2}\sin {\theta _2}\\ = \left[ {0.250 \times 1.50 \times \sin 55^\circ + 0.400 \times 896 \times \sin 312^\circ } \right]\;{\rm{kg}} \cdot {\rm{m/s}}\\ = 0\end{array}\)

Both are the same.

Most popular questions for Physics Textbooks

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of \(10.0kg\)and the horizontal component of its velocity is \(8.00 m/s\)when the \(65.0kg\)performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?

A \(3000\;{\rm{kg}}\)cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a \(15.0\;{\rm{kg}}\)shell at \(480\;{\rm{m/s}}\)at an angle of \(20.0^\circ \)above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired?

A \(5.50\;kg\) bowling ball moving at \(9.00\;{\rm{m/s}}\)collides with a \(0.850\;{\rm{kg}}\)bowling pin, which is scattered at an angle of \(85.{0^ \circ }\)to the initial direction of the bowling ball and with a speed of\(15.0\;{\rm{m/s}}\). (a) Calculate the final velocity (magnitude and direction) of the owling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.

Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum.

Two identical pucks collide on an air hockey table. One puck was originally at rest.

(a) If the incoming puck has a speed of \(6.00 m/s \)andscatters to an angle of ,what is the velocity (magnitude anddirection) of the second puck? (You may use the result that \({\theta _1}-{\theta _2} = 90°\) for elastic collisions of objects that have identicalmasses.)

(b) Confirm that the collision is elastic.
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