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Physics of Hearing

- Discuss two dimensional collisions as an extension of one dimensional analysis.
- Define point masses.
- Derive an expression for conservation of momentum along
*x*-axis and*y*-axis. - Describe elastic collisions of two objects with equal mass.
- Determine the magnitude and direction of the final velocity given initial velocity, and scattering angle.

One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of

We start by assuming that

Along the

*p*_{1x} + *p*_{2x} = *p′*_{1x} + *p′*_{2x}.

*m*_{1}*v*_{1x} + *m*_{2}*v*_{2x}* **= m*_{2}*v*_{2x} = *m*_{1}*v*′_{1x} + *m*_{2}*v*′_{2x}.

*m*_{1}*v*_{1}* _{x} *=

Conservation of momentum along the x-axis gives the following equation:

*m*_{1}*v*_{1} = *m*_{1}*v*′_{1} cos *θ*_{1} + *m*_{2}*v*′_{2} cos *θ*_{2}, where *θ*_{1} and *θ*_{2} are as shown in Figure 1.

*m*_{1}*v*_{1} = *m*_{1}*v*′_{1} cos *θ*_{1} + *m*_{2}*v*′_{2} cos *θ*_{2}

*p*_{1}* _{y} *+

0 = *m*_{1}*v*′_{1y}* *+ *m*_{2}*v*′_{2y}.

Thus, conservation of momentum along the

0 = *m*_{1}*v*′_{1} sin *θ*_{1} + *m*_{2}*v*′_{2} sin *θ*_{2}.

0 = *m*_{1}*v*′_{1} sin *θ*_{1} + *m*_{2}*v*′_{2} sin *θ*_{2}

The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (

Everything is known in these equations except

$\displaystyle\left(\tan\theta=\frac{\sin\theta}{\cos\theta}\right)\\$

, we obtain:$\displaystyle\tan\theta_{2}=\frac{v′_{1}\sin\theta_1}{v′_{1}\cos\theta_1-v_1}\\$

Entering known values into the previous equation gives$\tan{\theta }_{2}=\frac{\left(1\text{.}\text{50}\text{m/s}\right)\left(0\text{.}\text{7071}\right)}{\left(1\text{.}\text{50}\text{m/s}\right)\left(0\text{.}\text{7071}\right)-2\text{.}\text{00}\text{m/s}}=-1\text{.}\text{129}\\$

.
Thus,*θ*_{2 } = tan^{−1 } (−1.129) = 311.5º ≈ 312º.

$\displaystyle{v}′_{2}=\frac{m_1}{m_2}v′_{1}\frac{\sin\theta_{1}}{\sin\theta_{1}}\\$

Entering known values into this equation gives$\displaystyle{v}′_{2}=-\left(\frac{0.250\text{ kg}}{0.400\text{ kg}}\right)\left(1.50\text{ m/s}\right)\left(\frac{0.7071}{-0.7485}\right)\\$

Thus,*v′*_{2} = 0.886 m/s.

$\frac{1}{2}{mv_1}^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}\\$

.
Because the masses are equal, $\frac{1}{2}mv_1^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}+mv′_{1}v′_{2}\cos\left(\theta_{1}-\theta_{2}\right)\\$

.
(Remember that *mv*′_{1}*v*′_{2} cos(*θ*_{1} − *θ*_{2}) = 0.

*v*′_{1}= 0: head-on collision; incoming ball stops*v*′_{2}= 0: no collision; incoming ball continues unaffected- cos(
*θ*_{1}−*θ*_{2})=0: angle of separation (*θ*_{1}−*θ*_{2}) is 90º after the collision

All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called

- The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the
*x*-axis parallel to the velocity of the incoming particle. - Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the
*x*-axis), stated by*m*_{1}*v*_{1}=*m*_{1}*v*′_{1}cos*θ*_{1}+*m*_{2}*v*′_{2}cos*θ*_{2}and along the direction perpendicular to the initial direction (the*y*-axis) stated by 0 =*m*_{1}*v*′_{1y}+*m*_{2}*v*′_{2y}. - The internal kinetic before and after the collision of two objects that have equal masses is

$\frac{1}{2}{mv_{1}}^{2}=\frac{1}{2}{mv′_{1}}^{2}+\frac{1}{2}{mv′_{2}}^{2}+{\text{mv}′}_{1}{v′}_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)\\$

.
- Point masses are structureless particles that cannot spin.

- Figure 3 shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle
*θ*_{1}) at which the small object can emerge after colliding elastically with the cube. How does θ_{1}depend on*b*, the so-called impact parameter? Ignore any effects that might be due to rotation after the collision, and assume that the cube is much more massive than the small object. (b) Answer the same questions if the small object instead collides with a massive sphere.

- 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 and scatters to an angle of 30.0º, what is the velocity (magnitude and direction) of the second puck? (You may use the result that
*θ*_{1}−*θ*_{2}= 90º for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic. - Confirm that the results of the Example 1 do conserve momentum in both the
*x*- and*y*-directions. - A 3000-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-kg shell at 480 m/s at an angle of 20.0º 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?
**Professional Application.**A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0º to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling 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.**Professional Application.**Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (^{4}He) from gold-197 nuclei (^{197}Au). The energy of the incoming helium nucleus was 8.00 × 10^{−13}J, and the masses of the helium and gold nuclei were 6.68 × 10^{−27}kg and 3.29 × 10^{−25}kg, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120º during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?**Professional Application.**Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at 8.00m/s due south. The second car has a mass of 850 kg and is approaching at 17.0m/s due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x-axis and y-axis; instead, you must look for other simplifying aspects.- Starting with equations
*m*_{1}*v*_{1}=*m*_{1}*v*′_{1}cos*θ*_{1}+*m*_{2}*v*′_{2}cos*θ*_{2}and 0 =*m*_{1}*v*′_{1}sin*θ*_{1}+*m*_{2}*v*′_{2}sin*θ*_{2}for conservation of momentum in the*x*- and*y*-directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses,$\frac{1}{2}{{mv}_{1}}^{2}=\frac{1}{2}{{mv}′_{1}}^{2}+\frac{1}{2}{{mv}′_{2}}^{2}+{mv}′_{1}{v′}_{2}\cos\left({\theta }_{1}-{\theta }_{2}\right)\\$as discussed in the text. **Integrated Concepts.**A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away?

$\displaystyle{0}=m{v′ }_{1}\sin30^{\circ}-m{v′ }_{2}\sin60^{\circ}\Rightarrow {v′ }_{1}^{}={v′ }_{2}^{}\frac{\sin60^{\circ}}{\sin30^{\circ}}=\text{5.196 m/s}\\$

Verify that ratio of initial to final KE equals one:

$\begin{cases}\text{KE}=\frac{1}{2}{mv}_{1}^{2}=18m\text{ J}\\ \text{KE}=\frac{1}{2}{mv′_{1}}^{2}+\frac{1}{2}{mv′_{2}}^{2}=18m\text{ J}\end{cases}\\$

$\frac{\text{KE}}{\text{KE}}′=1.00\\$

3. (a) −2.26m/s; (b) 7.63 × 105. (a) 5.36 × 10

7. We are given that

${m}_{1}={m}_{2}\equiv m\\$

. The given equations then become:${v}_{1}={v}_{1}\text{cos}{\theta }_{1}+{v}_{2}\text{cos}{\theta }_{2}\\$

and

$0={v′}_{1}^{}\text{sin}{\theta }_{1}+{v′}_{2}^{}\text{sin}{\theta }_{2}\\$

.
Square each equation to get

$\begin{array}{lll}{{v}_{1}}^{2}& =& {{v′ }_{1}}^{2}{\text{cos}}^{2}{\theta }_{1}+{{v′ }_{2}}^{2}{\text{cos}}^{2}{\theta }_{2}+2{v′ }_{1}{v′}_{2}\text{cos}{\theta }_{1}\text{cos}{\theta }_{2}\\ 0& =& {{v′ }_{1}}^{2}{\text{sin}}^{2}{\theta }_{1}+{{v′ }_{2}}^{2}{\text{sin}}^{2}{\theta }_{2}+2{v′ }_{1}{v′ }_{2}\text{sin}{\theta }_{1}\text{sin}{\theta }_{2}\text{.}\end{array}\\$

Add these two equations and simplify:$\begin{array}{lll}{{v}_{1}}^{2}& =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{{v′ }_{1}}^{}{{v′ }_{2}}^{}\left(\text{cos}{\theta }_{1}\text{cos}{\theta }_{2}+\text{sin}{\theta }_{1}\text{sin}{\theta }_{2}\right)\\ & =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{v′ }_{1}{v′ }_{2}\left[\frac{1}{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)+\frac{1}{2}\text{cos}\left({\theta }_{1}+{\theta }_{2}\right)+\frac{1}{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)-\frac{1}{2}\text{cos}\left({\theta }_{1}+{\theta }_{2}\right)\right]\\ & =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{v′ }_{1}{v′ }_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right).\end{array}\\$

Multiply the entire equation by $\frac{1}{2}m\\$

to recover the kinetic energy:$\frac{1}{2}{{\mathit{\text{mv}}}_{1}}^{2}=\frac{1}{2}m{{v′ }_{1}}^{2}+\frac{1}{2}m{{v′ }_{2}}^{2}+m{v′ }_{1}{v′ }_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)\\$

as discussed in the text.