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
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.
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
$\text{30}\text{.}\mathrm{0\xba}$ ,what is the velocity (magnitude and direction) of the second puck? (You may use the result that
${\theta}_{1}-{\theta}_{2}=\text{90\xba}$ for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic.
(a) 3.00 m/s,
$\text{60\xba}$ below
$x$ -axis
(b) Find speed of first puck after collision:
$0=m{v\prime}_{1}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30\xba}-m{v\prime}_{2}^{}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\text{60\xba}\Rightarrow {v\prime}_{1}^{}={v\prime}_{2}^{}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{60\xba}}{\text{sin}\phantom{\rule{0.25em}{0ex}}\text{30\xba}}=\text{5.196 m/s}$
Verify that ratio of initial to final KE equals one:
$\left(\begin{array}{l}\text{KE}=\frac{1}{2}{{\mathrm{mv}}_{1}}^{2}=18m\phantom{\rule{0.25em}{0ex}}\text{J}\\ \text{KE}=\frac{1}{2}{{\mathrm{mv}\prime}_{1}}^{2}+\frac{1}{2}{{\mathrm{mv}\prime}_{2}}^{2}=18m\phantom{\rule{0.25em}{0ex}}\text{J}\end{array}\right\}\frac{\text{KE}}{\text{KE\u2032}}=1.00$
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
$\text{20}\text{.}\mathrm{0\xba}$ 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?
(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.
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
$\text{85}\text{.}\mathrm{0\xba}$ 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.
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
$8.00\times {\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , and the masses of the helium and gold nuclei were
$6.68\times {\text{10}}^{-\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and
$3.29\times {\text{10}}^{-\text{25}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of
$\text{120\xba}$ 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)
$5\text{.}\text{36}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ at
$-\text{29.5\xba}$
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,
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?
you shouldn't say distance,displacement and distance are two different things .distance can be lopped curved but displacement is always in a straight line so you can't use distance to define it. displacement is the change of position in a specified direction.
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g
Smaller mass=2g
Momentum of both=8
Meaning V for L =2 and V for S=4
Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the
initial speed of the other. If the faster stone takes 10 s to return to the ground, how
long will it take the slower stone to return? If the slower stone reaches a maximum
height of H, how high will the faster stone go
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.