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?
A few grains of table salt were put in a cup of cold water kept at constant temperature and left undisturbed. eventually all the water tasted salty. this is due to?
20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K)
Formula.
(ms*cs+C)*T
20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K)
Formula.
(ms*cs+C)*T
Lilian
because it changes only direction and the speed is kept constant
It's filtered light from the 2 forms of radiation emitted from the sun. It's mainly filtered UV rays. There's a theory titled Scatter Theory that covers this topic
Mike
A heating coil of resistance 30π is connected to a 240v supply for 5min to boil a quantity of water in a vessel of heat capacity 200jk. If the initial temperature of water is 20°c and it specific heat capacity is 4200jkgk calculate the mass of water in a vessel
A thin equi convex lens is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens concise in position with its own image the space between the undersurface of d lens and the mirror is filled with water (refractive index =1•33)and then to concise with d image d pin has to
A monkey throws a coconut straight upwards from a coconut tree with a velocity of 10 ms-1. The coconut tree is 30 m high. Calculate the maximum height of the coconut from the top of the coconut tree? Can someone answer my question
Wheatstone bridge is an instrument used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
MUHD
Rockwell Software is Rockwell Automation’s "Retro Encabulator".
Now, basically the only new principle involved is that instead of power being generated by the relative motion of conductors and fluxes, it’s produced by the modial interaction of magneto-reluctance and capacitive diractance. The origin