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Q100P

Expert-verifiedFound in: Page 255

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at 3.50 m/s along a line making an angle of${\mathbf{22}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\xb0}}$ 2 with the cue ball’s original direction of motion, and the second ball has a speed of ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$ . Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don’t consider the rotation) conserved?**

- The angle between the direction of motion of the second ball and the original direction of motion of the cue ball $\varphi \mathrm{is}41.0\xb0$ .
- The original speed of the cue ball ${v}_{i1}$is $4.75\mathrm{m}/\mathrm{s}$ .
- Kinetic energy is not conserved.

- The final speed of the cue ball, ${v}_{f1}\mathrm{is}3.50\mathrm{m}/\mathrm{s}$ .
- After collision angle of the cue ball, $\theta \mathrm{is}22\xb0$ .
- The initial speed of the second ball, ${V}_{i2}\mathrm{is}0\mathrm{m}/\mathrm{s}$ .
- The final speed of the second ball, ${V}_{f2}\mathrm{is}2.00\mathrm{m}/\mathrm{s}$ .

**You can use the concept of conservation of momentum and conservation of kinetic energy. You can find the angle for the second ball after the collision for the first part by writing the equations of conservation on momentum for the x and y directions and solving these equations. Also, from the equation of conservation of momentum in the horizontal direction, you can find the original speed of the cue ball. Using the conservation of kinetic energy equation, you can determine whether kinetic energy is conserved or not.**

**${m}_{1}{v}_{i1}+{m}_{2}{v}_{i2}={m}_{1}{v}_{f1}+{m}_{2}{v}_{f2}\phantom{\rule{0ex}{0ex}}\left(\frac{1}{2}\right){m}_{1}{v}_{i1}^{2}+\left(\frac{1}{2}\right){m}_{1}{v}_{i2}^{2}+\left(\frac{1}{2}\right){m}_{1}{v}_{f1}^{2}+\left(\frac{1}{2}\right){m}_{2}{v}_{f2}^{2}$**

We can write two equations of conservation of momentum for x and y directions.

** **

For x direction, we can write,

${m}_{1}{v}_{i1}+{m}_{2}{v}_{i2}={m}_{1}{v}_{f1}\mathrm{cos}\theta +{m}_{2}{v}_{f2}\mathrm{cos}\varphi $

Where $\varphi $ is the angle of the second ball after the collision, and are the initial velocities of the balls, and masses are the same for both the balls so that we can cancel them as,

${v}_{i1}+{v}_{i2}={v}_{f1}\mathrm{cos}\theta +{v}_{f2}\mathrm{cos}\varphi $

Initially, the second ball is at rest; we can write,

${v}_{i1}+0={v}_{f1}\mathrm{cos}\left(22\xb0\right)+\left(2.00\right)\mathrm{cos}\varphi \phantom{\rule{0ex}{0ex}}{v}_{i1}={v}_{f1}\mathrm{cos}\mathrm{cos}\left(22\xb0\right)+\left(2.00\right)\mathrm{cos}\varphi $ (1)

Similarly, we can write for u direction,

${v}_{i1}+{v}_{i2}={v}_{f1}\mathrm{sin}\theta +{v}_{f2}\mathrm{sin}\varphi \phantom{\rule{0ex}{0ex}}0={v}_{f1}\mathrm{sin}\theta +{v}_{f2}\mathrm{sin}\varphi \phantom{\rule{0ex}{0ex}}{v}_{f1}\mathrm{sin}(22\xb0)={v}_{f2}\mathrm{sin}\varphi $ (2)

Substitute the values in the above expression, and we get,

$\left(3.50\right)\mathrm{sin}\left(22\xb0\right)=\left(2.00\right)\mathrm{sin}\varphi \phantom{\rule{0ex}{0ex}}\mathrm{sin}\varphi =\frac{3.50}{2.00}\mathrm{sin}22\xb0\phantom{\rule{0ex}{0ex}}\varphi ={\mathrm{sin}}^{-1}\left(0.6555\right)$

Solving further as,

$\varphi =\left(0.6555\right)\phantom{\rule{0ex}{0ex}}=40.96\xb0\phantom{\rule{0ex}{0ex}}=41.0\xb0$

Thus, the angle between the direction of motion of the second ball and the original direction of motion of the cue ball $\varphi \mathrm{is}41.0\xb0$ .

The original speed of the cue ball**:**

Using the angle found in part a) in equation (1), we can find the original velocity speed of the cue ball. We can write,

${v}_{i1}=\left(3.50\right)\mathrm{cos}\left(22\xb0\right)+\left(2.00\right)\mathrm{cos}\left(41.0\xb0\right)\phantom{\rule{0ex}{0ex}}=4.75\mathrm{m}/\mathrm{s}$

Thus, the original speed of the cue ball ${V}_{i1}\mathrm{is}4.75\mathrm{m}/\mathrm{s}$${v}_{i1}\mathrm{is}4.75\mathrm{m}/\mathrm{s}$ .

We can calculate initial and final kinetic energies separately to determine whether kinetic energy is conserved or not. We can write,

$K.{E}_{I}=\left(\frac{1}{2}\right){m}_{1}{v}_{i1}^{2}+\left(\frac{1}{2}\right){m}_{1}{v}_{i2}^{2}$

Substitute the values in the above expression, and we get,

$\mathrm{K}.{\mathrm{E}}_{\mathrm{I}}=0.5\mathrm{m}{\left(4.75\right)}^{2}+0.5\mathrm{m}\left(0\right)\phantom{\rule{0ex}{0ex}}=11.28\mathrm{m}\phantom{\rule{0ex}{0ex}}=11.3\mathrm{m}$

Similarly, we can write for final kinetic energy,

$K.{E}_{I}=\left(\frac{1}{2}\right){m}_{1}{v}_{f1}^{2}+\left(\frac{1}{2}\right){m}_{1}{v}_{f2}^{2}$

Substitute the values in the above expression, and we get,

$K.{E}_{f}=0.5m{\left(3.50\right)}^{2}+0.5\left(m\right){(2.00)}^{2}\phantom{\rule{0ex}{0ex}}=6.125m+2m\phantom{\rule{0ex}{0ex}}=8.125m\phantom{\rule{0ex}{0ex}}=8.1m$

We get different values for initial and final kinetic energies so that kinetic energy is not conserved.

Thus, kinetic energy is not conserved.

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