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9.4 Two-Dimensional CollisionsPowerPoint Presentation

9.4 Two-Dimensional Collisions

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9.4 Two-Dimensional Collisions

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- The momentum is conserved in all directions
- Use subscripts for
- identifying the object
- indicating initial or final values
- the velocity components

- If the collision is elastic, use conservation of kinetic energy as a second equation
- Remember, the simpler equation can only be used for one-dimensional situations

Qualitative Analysis

- Physical Principles:The same as in One-Dimension
- 1. Conservation of VECTOR momentum:
P1x + P2x = P1x + P2x &P1y + P2y= P1y + P2y

- 2. Conservation of Kinetic Energy
½m1v12 + ½m2v22 = ½m1v’12 + ½m2v’22

- For a collision of two particles in two dimensions implies that the momentum in each direction xand yis conserved
- The game of billiards is an example for such two dimensional collisions
- The equations for conservation of momentum are:
m1v1ix+m2v2ix ≡m1v1fx+m2v2fx

m1v1iy+m2v2iy ≡m1v1fy+m2v2fy

- Subscripts represent:
- (1,2) Objects
- (i,f) Initial and final values
- (x,y) Component direction

- Particle 1 is moving at velocity v1iand particle 2 is at rest
- In the x-direction, the initial momentum is m1v1i
- In the y-direction, the initial momentum is 0

- After the glancing collision, the conservation of momentum in the x-direction is
m1v1i≡m1v1fcosq+m2v2f cosf (9.24)

- After the collision, the conservation of momentum in the y-direction is
0 ≡ m1v1fsinq+m2v2f sinf(9.25)

- Mass of the car mc= 1500kg
- Mass of the van mv = 2500kg
- Findvfif this is a perfectly inelastic collision (they stick together).
- Before collision
- The car’smomentum is:
Σpxi= mcvc

Σpxi= (1500)(25) = 3.75x104 kg·m/s

- The van’smomentum is:
Σpyi= mvvv

Σpyi= (2500)(20) = 5.00x104 kg·m/s

- The car’smomentum is:

- After collision,both have the same x- and y-components:
Σpxf = (mc + mv )vf cos

Σpyf = (mc + mv )vf sin

- Because the total momentum is both directions is conserved:
Σpxf = Σpxi

3.75x104 kg·m/s = (mc + mv )vf cos (1)

Σpyf = Σpyi

5.00x104 kg·m/s = (mc + mv )vf sin (2)

- Since (mc + mv ) = 400kg.
3.75x104 kg·m/s = 4000vf cos (1)

5.00x104 kg·m/s = 4000vf sin (2)

- Dividing Eqn (2) by (1)
5.00/3.75 =1.33 = tan

= 53.1°

- Substitutingin Eqn (2) or (1)
5.00x104 kg·m/s = 4000vf sin53.1°

vf =5.00x104/(4000sin53.1° )

vf =15.6m/s

- There is a special point in a system or object, called the center of mass (CM), that moves as if all of the mass of the system is concentrated at that point
- The system will move as if an external force were applied to a single particle of mass M located at the CM
- M= Σmiis the total mass of the system

- The coordinates of the center of mass are
(9.28) (9.29)

- The center of mass can be located by its position vector, rCM
(9.30)

- ri is the position of the i th particle, defined by

- Both masses are on the x-axis
- The center of mass (CM) is on the x-axis
- One dimension, x-axis
xCM = (m1x1 + m2x2)/M

M = m1+m2

xCM≡(m1x1 + m2x2)/(m1+m2)

- The center of mass is closer to the particle with the larger mass
xCM≡(m1x1 + m2x2)/(m1+m2)

- If: x1 = 0, x2 = d & m2 = 2m1
xCM≡(0 + 2m1d)/(m1+2m1)

xCM≡2m1d/3m1

xCM = 2d/3

- Up to now, we’ve been mainly concerned with the motion of single (point) particles.
- To treat extended bodies, we’ve approximated the body as a point particle & treated it as if it had all of its mass at a point!
How is this possible?

- Real, extended bodies have complex motion, including: translation, rotation, & vibration!
- Think of the extended object as a system containing a large number of particles

- The particle separationis very small, so the mass can be considered a continuous mass distribution:
- The coordinates of theCM of the object are:
(9.31)

(9.32)

- The position of CM can also be found by:
(9.33)

- The CM of any symmetrical object lies on an axis of symmetry and on any plane of symmetry
- An extended object can be considered a distribution of small mass elements, Dm
- The CM is located at position rCM

A group of extended bodies, each with a known CM and

equivalent mass m.

Find the CM of the group.

xCM = (Σmixi)/Σmi

xCM = (mx1+ mx2+ mx3)/(m+m+m)

xCM = m(x1+ x2+ x3)/3m = (x1+ x2+ x3)/3

xCM = (1.00m + 5.00m + 6.00m)/3 = 4.00m

- Find the CM position of a rod of mass M and length L
- The location is on the x-axis
(yCM = zCM = 0)

- (A). Assuming the road has a uniform mass per unit length
λ = M/L (Linear mass density)

- From Eqn 9.31

- But λ = M/L
- (B). Assuming now that the linear mass density of the road is no uniform: λ = x
- The CM will be:

- But mass of the rod and are related by:
The CM will be:

Material for the Final

- Examples to Read!!!
- Example 9.12(page 269)
- Example 9.13(page 272)
- Example 9.16(page 276)

- Homework to be solved in Class!!!
- Problems: 43