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Expert-verified Found in: Page 292 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # 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 $7.11×{10}^{-3}J$ and $2.84×{10}^{-2}J$.

See the step by step solution

## 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:

${v}_{1}=\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right){u}_{1}+\left(\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}\right){u}_{2}$

Substitute all the values in the above equation.

${v}_{1}=\left(\frac{2g-1g}{2g+1g}\right)\left(8m/s\right)+\left(\frac{2\left(1g\right)}{2g+1g}\right)\left(0m/s\right)\phantom{\rule{0ex}{0ex}}{v}_{1}=2.67m/s$

The velocity of stationaryparticle after collision is given as:

${v}_{2}=\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right){u}_{2}+\left(\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}\right){u}_{1}$

Substitute all the values in the above equation.

${v}_{2}=\left(\frac{1g-2g}{2g+1g}\right)\left(0m/s\right)+\left(\frac{2\left(2g\right)}{2g+1g}\right)\left(8m/s\right)\phantom{\rule{0ex}{0ex}}{v}_{2}=10.7m/s\phantom{\rule{0ex}{0ex}}$

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:

${v}_{1}=\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right){u}_{1}+\left(\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}\right){u}_{2}$

Substitute all the values in the above equation.

${v}_{1}=\left(\frac{2g-10g}{2g+10g}\right)\left(8m/s\right)+\left(\frac{2\left(10g\right)}{2g+1g}\right)\left(0m/s\right)\phantom{\rule{0ex}{0ex}}{v}_{1}=-5.33m/s$

The velocity of stationary particle after collision is given as:

${v}_{2}=\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right){u}_{2}+\left(\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}\right){u}_{1}$

Substitute all the values in the above equation.

${v}_{2}=\left(\frac{10g-2g}{2g+10g}\right)\left(0m/s\right)+\left(\frac{2\left(2g\right)}{2g+10g}\right)\left(8m/s\right)\phantom{\rule{0ex}{0ex}}{v}_{2}=2.67m/s$

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.

${K}_{a}=\frac{1}{2}\left(2g\right)\left(\frac{1kg}{1000g}\right){\left(2.67m/s\right)}^{2}\phantom{\rule{0ex}{0ex}}{K}_{a}=7.11×{10}^{-3}J$

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.

${K}_{a}=\frac{1}{2}\left(2g\right)\left(\frac{1kg}{1000g}\right){\left(-5.33m/s\right)}^{2}\phantom{\rule{0ex}{0ex}}{K}_{a}=2.84×{10}^{-2}J$

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 $7.11×{10}^{-3}J$ and $2.84×{10}^{-2}J$ . ### Want to see more solutions like these? 