# 8.6 Collisions of point masses in two dimensions  (Page 2/5)

 Page 2 / 5

But because particle 2 is initially at rest, this equation becomes

${m}_{1}{v}_{1x}={m}_{1}{v\prime }_{1x}^{}+{m}_{2}{v\prime }_{2x}^{}.$

The components of the velocities along the $x$ -axis have the form $v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ . Because particle 1 initially moves along the $x$ -axis, we find ${v}_{1x}={v}_{1}$ .

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

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

where ${\theta }_{1}$ and ${\theta }_{2}$ are as shown in [link] .

## Conservation of momentum along the $x$ -axis

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

Along the $y$ -axis, the equation for conservation of momentum is

${p}_{1y}+{p}_{2y}={p\prime }_{1y}^{}+{p\prime }_{2y}^{}$

or

${m}_{1}{v}_{1y}+{m}_{2}{v}_{2y}={m}_{1}{v\prime }_{1y}^{}+{m}_{2}{v\prime }_{2y}^{}.$

But ${v}_{1y}$ is zero, because particle 1 initially moves along the $x$ -axis. Because particle 2 is initially at rest, ${v}_{2y}$ is also zero. The equation for conservation of momentum along the $y$ -axis becomes

$0={m}_{1}{v\prime }_{1y}^{}+{m}_{2}{v\prime }_{2y}^{}.$

The components of the velocities along the $y$ -axis have the form $v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ .

Thus, conservation of momentum along the $y$ -axis gives the following equation:

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

## Conservation of momentum along the $y$ -axis

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

The equations of conservation of momentum along the $x$ -axis and $y$ -axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.

## Determining the final velocity of an unseen object from the scattering of another object

Suppose the following experiment is performed. A 0.250-kg object $\left({m}_{1}\right)$ is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg $\left({m}_{2}\right)$ . The 0.250-kg object emerges from the room at an angle of $\text{45}\text{.}0º$ with its incoming direction.

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 $\left({v\prime }_{2}^{}$ and ${\theta }_{2}\right)$ of the 0.400-kg object after the collision.

Strategy

Momentum is conserved because the surface is frictionless. The coordinate system shown in [link] is one in which ${m}_{2}$ is originally at rest and the initial velocity is parallel to the $x$ -axis, so that conservation of momentum along the $x$ - and $y$ -axes is applicable.

Everything is known in these equations except ${v\prime }_{2}^{}$ and ${\theta }_{2}$ , which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the $x$ - and $y$ -directions.

Solution

Solving ${m}_{1}{v}_{1}={m}_{1}{v\prime }_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}+{m}_{2}{v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ for ${v}_{2}^{\prime }\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ 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 ${v\prime }_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}$ and taking the ratio yields an equation (in which θ 2 is the only unknown quantity. Applying the identity $\left(\text{tan}\phantom{\rule{0.25em}{0ex}}\theta =\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }{\text{cos}\phantom{\rule{0.25em}{0ex}}\theta }\right)$ , we obtain:

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

Entering known values into the previous equation gives

$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}=\frac{\left(1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(0\text{.}\text{7071}\right)}{\left(1\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(0\text{.}\text{7071}\right)-2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=-1\text{.}\text{129}.$

Thus,

${\theta }_{2}={\text{tan}}^{-1}\left(-1\text{.}\text{129}\right)=\text{311}\text{.}5º\approx \text{312º}.$

Angles are defined as positive in the counter clockwise direction, so this angle indicates that ${m}_{2}$ is scattered to the right in [link] , as expected (this angle is in the fourth quadrant). Either equation for the $x$ - or $y$ -axis can now be used to solve for ${v\prime }_{2}$ , but the latter equation is easiest because it has fewer terms.

how many subject is in physics
the write question should be " How many Topics are in O- Level Physics, or other branches of physics.
effiom
how many topic are in physics
Praise
yh I need someone to explain something im tryna solve . I'll send the question if u down for it
a ripple tank experiment a vibrating plane is used to generate wrinkles in the water .if the distance between two successive point is 3.5cm and the wave travel a distance of 31.5cm find the frequency of the vibration
Tamdy
the range of objects and phenomena studied in physics is
what is Linear motion
straight line motion is called linear motion
then what
Amera
linear motion is a motion in a line, be it in a straight line or in a non straight line. It is the rate of change of distance.
Saeedul
Hi
aliyu
Richard
Linear motion is a one-dimensional motion along a straight line, and can therefore be described mathematically using only one spatial dimension
Jason
is a one-dimensional motion along a straight line, and can therefore be described mathematically using only one spatial dimensions.
Praise
what is a classical electrodynamics?
Marga
what is dynamics
Marga
dynamic is the force that stimulates change or progress within the system or process
Oze
what is the formula to calculate wavelength of the incident light
if a spring is is stiffness of 950nm-1 what work will be done in extending the spring by 60mmp
State the forms of energy
machanical
Ridwan
Word : Mechanical wave Definition : The waves, which need a material medium for their propagation, e.g., Sound waves. \n\nOther Definition: The waves, which need a material medium for their propagation, are called mechanical waves. Mechanical waves are also called elastic waves. Sound waves, water waves are examples of mechanical waves.t Definition: wave consisting of periodic motion of matter; e.g. sound wave or water wave as opposed to electromagnetic wave.h
correct
Akinpelu
what is mechanical wave
a wave which require material medium for its propagation
syed
The S.I unit for power is what?
watt
Okoli
Am I correct
Okoli
it can be in kilowatt, megawatt and so
Femi
yes
Femi
correct
Jaheim
kW
Akinpelu
OK that's right
Samuel
SI.unit of power is.watt=j/c.but kw.and Mw are bigger.umots
syed
What is physics
study of matter and its nature
Akinpelu
The word physics comes from a Greek word Physicos which means Nature.The Knowledge of Nature. It is branch of science which deals with the matter and energy and interaction between them.
Uniform
why in circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction
reasonable
Femi
because it is balanced by the inward acceleration otherwise known as centripetal acceleration
MUSTAPHA
What is a wave
Tramsmission of energy through a media
Mateo
is the disturbance that carry materials as propagation from one medium to another
Akinpelu
mistakes thanks
Akinpelu
find the triple product of (A*B).C given that A =i + 4j, B=2i - 3j and C = i + k
Difference between north seeking pole and south seeking pole