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  4. 6.7 collisions of point masses

0.6 6.7 collisions of point masses in two dimensions  (Page 4/5)

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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 ′ 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 ′ 1 y + m 2 v ′ 2 y .
  • The internal kinetic before and after the collision of two objects that have equal masses is
    1 2 mv 1 2 = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 + mv ′ 1 v ′ 2 cos θ 1 − θ 2 .
  • Point masses are structureless particles that cannot spin.

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 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.

(a) 3.00 m/s, 60º below x size 12{x} {} -axis

(b) Find speed of first puck after collision: 0 = m v ′ 1 sin 30º − m v ′ 2 sin 60º ⇒ v ′ 1 = v ′ 2 sin 60º sin 30º = 5.196 m/s

Verify that ratio of initial to final KE equals one: KE = 1 2 mv 1 2 = 18 m J KE = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 = 18 m J KE KE′ = 1.00

Confirm that the results of the example [link] do conserve momentum in both the x size 12{x} {} - and y size 12{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 size 12{"20" "." 0°} {} 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 . 26 m/s size 12{ - 2 "." "26"`"m/s"} {}

(b) 7 . 63 × 10 3 J size 12{7 "." "63" times "10" rSup { size 8{3} } `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

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?

(a) 5 . 36 × 10 5 m/s at − 29.5º

(b) 7 . 52 × 10 − 13 J size 12{7 "." "52" times "10" rSup { size 8{ - "13"} } `J} {}

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,

1 2 mv 1 2 = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 + mv ′ 1 v ′ 2 cos θ 1 − θ 2

as discussed in the text.

We are given that m 1 = m 2 ≡ m size 12{m rSub { size 8{1} } =m rSub { size 8{2} } equiv m} {} . The given equations then become:

v 1 = v 1 cos θ 1 + v 2 cos θ 2

and

0 = v ′ 1 sin θ 1 + v ′ 2 sin θ 2 .

Square each equation to get

v 1 2 = v ′ 1 2 cos 2 θ 1 + v ′ 2 2 cos 2 θ 2 + 2 v ′ 1 v ′ 2 cos θ 1 cos θ 2 0 = v ′ 1 2 sin 2 θ 1 + v ′ 2 2 sin 2 θ 2 + 2 v ′ 1 v ′ 2 sin θ 1 sin θ 2 .

Add these two equations and simplify:

v 1 2 = v ′ 1 2 + v ′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 cos θ 2 + sin θ 1 sin θ 2 = v ′ 1 2 + v ′ 2 2 + 2 v ′ 1 v ′ 2 1 2 cos θ 1 − θ 2 + 1 2 cos θ 1 + θ 2 + 1 2 cos θ 1 − θ 2 − 1 2 cos θ 1 + θ 2 = v ′ 1 2 + v ′ 2 2 + 2 v ′ 1 v ′ 2 cos θ 1 − θ 2 .

Multiply the entire equation by 1 2 m size 12{ { { size 8{1} } over { size 8{2} } } m} {} to recover the kinetic energy:

1 2 mv 1 2 = 1 2 m v ′ 1 2 + 1 2 m v ′ 2 2 + m v ′ 1 v ′ 2 cos θ 1 − θ 2

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

what does the ideal gas law states
Joy Reply
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
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Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
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Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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Source:  OpenStax, Unit 6 - momentum. OpenStax CNX. Jan 22, 2016 Download for free at https://legacy.cnx.org/content/col11961/1.1
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