# 8.6 Collisions of point masses in two dimensions  (Page 4/5)

 Page 4 / 5

## Section summary

• 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 ${m}_{1}{v}_{1}={m}_{1}{v\prime }_{1}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{m}_{2}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ and along the direction perpendicular to the initial direction (the $y$ -axis) stated by $0={m}_{1}{v\prime }_{1y}+{m}_{2}{v\prime }_{2y}$ .
• The internal kinetic before and after the collision of two objects that have equal masses is
$\frac{1}{2}{{\text{mv}}_{1}}^{2}=\frac{1}{2}{{\text{mv}\prime }_{1}}^{2}+\frac{1}{2}{{\text{mv}\prime }_{2}}^{2}+{\text{mv}\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right).$
• Point masses are structureless particles that cannot spin.

## Conceptual questions

[link] shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle ${\theta }_{1}$ ) at which the small object can emerge after colliding elastically with the cube. How does ${\theta }_{1}$ 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.

## Problems&Exercises

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 $\text{30}\text{.}0º$ ,what is the velocity (magnitude and direction) of the second puck? (You may use the result that ${\theta }_{1}-{\theta }_{2}=\text{90º}$ for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic.

(a) 3.00 m/s, $\text{60º}$ below $x$ -axis

(b) Find speed of first puck after collision: $0=m{v\prime }_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30º}-m{v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\text{60º}⇒{v\prime }_{1}^{}={v\prime }_{2}^{}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{60º}}{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30º}}=\text{5.196 m/s}$

Verify that ratio of initial to final KE equals one: $\left(\begin{array}{l}\text{KE}=\frac{1}{2}{{\mathrm{mv}}_{1}}^{2}=18m\phantom{\rule{0.25em}{0ex}}\text{J}\\ \text{KE}=\frac{1}{2}{{\mathrm{mv}\prime }_{1}}^{2}+\frac{1}{2}{{\mathrm{mv}\prime }_{2}}^{2}=18m\phantom{\rule{0.25em}{0ex}}\text{J}\end{array}}\frac{\text{KE}}{\text{KE′}}=1.00$

Confirm that the results of the example [link] 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 $\text{20}\text{.}0º$ 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?

(a) $-2\text{.}\text{26}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

(b) $7\text{.}\text{63}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J}$

(c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots.

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 $\text{85}\text{.}0º$ 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 $\left({}^{4}\text{He}\right)$ from gold-197 nuclei $\left({}^{\text{197}}\text{Au}\right)$ . The energy of the incoming helium nucleus was $8.00×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , and the masses of the helium and gold nuclei were $6.68×{\text{10}}^{-\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and $3.29×{\text{10}}^{-\text{25}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of $\text{120º}$ 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?

(a) $5\text{.}\text{36}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ at $-\text{29.5º}$

(b) $7\text{.}\text{52}×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}$

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 $8\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ due south. The second car has a mass of 850 kg and is approaching at $\text{17}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}$ 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 ${m}_{1}{v}_{1}={m}_{1}{v\prime }_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{m}_{2}{v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ and $0={m}_{1}{v\prime }_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{m}_{2}{v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ 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,

$\frac{1}{2}{{\text{mv}}_{1}}^{2}=\frac{1}{2}{{\text{mv}\prime }_{1}}^{2}+\frac{1}{2}{{\text{mv}\prime }_{2}}^{2}+{\text{mv}\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}-{\theta }_{2}\right)$

as discussed in the text.

We are given that ${m}_{1}={m}_{2}\equiv m$ . The given equations then become:

${v}_{1}={v}_{1}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{v}_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$

and

$0={v\prime }_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}.$

Square each equation to get

$\begin{array}{lll}{{v}_{1}}^{2}& =& {{v\prime }_{1}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{{v\prime }_{2}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}+2{v\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}\\ 0& =& {{v\prime }_{1}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{sin}}^{2}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{{v\prime }_{2}}^{2}\phantom{\rule{0.25em}{0ex}}{\text{sin}}^{2}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}+2{v\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}\text{.}\end{array}$

Add these two equations and simplify:

$\begin{array}{lll}{{v}_{1}}^{2}& =& {{v\prime }_{1}}^{2}+{{v\prime }_{2}}^{2}+2{{v\prime }_{1}}^{}{{v\prime }_{2}}^{}\left(\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}+\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}\right)\\ & =& {{v\prime }_{1}}^{2}+{{v\prime }_{2}}^{2}+2{v\prime }_{1}{v\prime }_{2}\left[\frac{1}{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}-{\theta }_{2}\right)+\frac{1}{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}+{\theta }_{2}\right)+\frac{1}{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}-{\theta }_{2}\right)-\frac{1}{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}+{\theta }_{2}\right)\right]\\ & =& {{v\prime }_{1}}^{2}+{{v\prime }_{2}}^{2}+2{v\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left({\theta }_{1}-{\theta }_{2}\right).\end{array}$

Multiply the entire equation by $\frac{1}{2}m$ to recover the kinetic energy:

$\frac{1}{2}{{\mathit{\text{mv}}}_{1}}^{2}=\frac{1}{2}m{{v\prime }_{1}}^{2}+\frac{1}{2}m{{v\prime }_{2}}^{2}+m{v\prime }_{1}{v\prime }_{2}\phantom{\rule{0.25em}{0ex}}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)$

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?

#### Questions & Answers

Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
which topic u discussing about?
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
hi
Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
hi guys
thomas
the meaning of phrase in physics
is the meaning of phrase in physics
Chovwe
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
how does a model differ from a theory
To use the vocabulary of model theory and meta-logic, a theory is a set of sentences which can be derived from a formal model using some rule of inference (usually just modus ponens). So, for example, Number Theory is the set of sentences true about numbers. But the model is a structure together wit
Jesilda
with an iterpretation.
Jesilda