Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

A 2.00-g particle moving at 8.00 m/s makes a perfectly elastic head-on collision with a resting 1.00-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10.0 g. (c) Find the final kinetic energy of the incident 2.00-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

(a) The speed of moving and stationary particle after collision is 2.67 m/s and 10.7 m/s.

(b) The speed of moving and stationary particle after collision is -5.33 m and 2.67 m/s.

(c) The kinetic energy of moving particle for case (a) and (b) are $7.11 \times 10^{- 3} J$ and $2.84 \times 10^{- 2} J$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of given data

The mass of moving particle is m₁= 2 g

The speed of moving particle is u₁ = 8 m/s

The mass of stationary particle is m₂= 1 g

The velocity of stationary particle is u₂ = 0 m/s

The replaced mass of stationary particle is m₂= 10 g

Step 2: Understanding the concept

The center of mass of system of particles is the point at which total mass of the particle can be assumed to be act.

Step 3: Determination of speed of each particle after collision

(a)

The speed of movingparticle after collision is given as:

$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + (\frac{2 m_{2}}{m_{1} + m_{2}}) u_{2}$

Substitute all the values in the above equation.

$v_{1} = (\frac{2 g - 1 g}{2 g + 1 g}) (8 m / s) + (\frac{2 (1 g)}{2 g + 1 g}) (0 m / s) v_{1} = 2.67 m / s$

The velocity of stationaryparticle after collision is given as:

$v_{2} = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) u_{2} + (\frac{2 m_{1}}{m_{1} + m_{2}}) u_{1}$

Substitute all the values in the above equation.

$v_{2} = (\frac{1 g - 2 g}{2 g + 1 g}) (0 m / s) + (\frac{2 (2 g)}{2 g + 1 g}) (8 m / s) v_{2} = 10.7 m / s$

Therefore, the speed of moving and stationaryparticle after collision is 2.67 m/s and 10.7 m/s.

Step 4: Determination of speed of each particle after collision

(b)

The speed of moving particle after collision is given as:

$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + (\frac{2 m_{2}}{m_{1} + m_{2}}) u_{2}$

Substitute all the values in the above equation.

$v_{1} = (\frac{2 g - 10 g}{2 g + 10 g}) (8 m / s) + (\frac{2 (10 g)}{2 g + 1 g}) (0 m / s) v_{1} = - 5.33 m / s$

The velocity of stationary particle after collision is given as:

$v_{2} = (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) u_{2} + (\frac{2 m_{1}}{m_{1} + m_{2}}) u_{1}$

Substitute all the values in the above equation.

$v_{2} = (\frac{10 g - 2 g}{2 g + 10 g}) (0 m / s) + (\frac{2 (2 g)}{2 g + 10 g}) (8 m / s) v_{2} = 2.67 m / s$

Therefore, the speed of moving and stationary particle after collision is -5.33 m/s and 2.67 m/s.

Step 5: Determination of final kinetic energy of moving particle after collision

(c)

The final kinetic energy of moving particle after collision in case (a) is given as:

$K_{a} = \frac{1}{2} m_{1} v_{1}^{2}$

Substitute all the values in the above equation.

$K_{a} = \frac{1}{2} (2 g) (\frac{1 k g}{1000 g}) {(2.67 m / s)}^{2} K_{a} = 7.11 \times 10^{- 3} J$

The final kinetic energy of moving particle after collision in case (b) is given as:

$K_{b} = \frac{1}{2} m_{1} v_{1}^{2}$

Substitute all the values in the above equation.

$K_{a} = \frac{1}{2} (2 g) (\frac{1 k g}{1000 g}) {(- 5.33 m / s)}^{2} K_{a} = 2.84 \times 10^{- 2} J$

The lost kinetic energy of moving particle for case (a) is more than case (b).

Therefore, the kinetic energy of moving particle for case (a) and (b) are $7.11 \times 10^{- 3} J$ and $2.84 \times 10^{- 2} J$ .

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Identification of given data

Step 2: Understanding the concept

Step 3: Determination of speed of each particle after collision

Step 4: Determination of speed of each particle after collision

Step 5: Determination of final kinetic energy of moving particle after collision

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Identification of given data

Step 2: Understanding the concept

Step 3: Determination of speed of each particle after collision

Step 4: Determination of speed of each particle after collision

Step 5: Determination of final kinetic energy of moving particle after collision

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