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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?
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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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