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91 P

Expert-verifiedFound in: Page 292

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 $7.11\times {10}^{-3}J$ and $2.84\times {10}^{-2}J$.

The mass of moving particle is m_{1 }= 2 g

The speed of moving particle is u_{1} = 8 m/s

The mass of stationary particle is m_{2}_{ }= 1 g

The velocity of stationary particle is u_{2} = 0 m/s

The replaced mass of stationary particle is m_{2}_{ }= 10 g

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

**(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.

**(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.

**(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.

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

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.

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

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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