Glancing Collisions In 2D, help simplifying

Click For Summary
SUMMARY

The forum discussion focuses on a physics problem involving glancing collisions in 2D, specifically a scenario where mass m1 collides with mass 2m at rest. The resulting velocities V1 and V2 are calculated using conservation of momentum equations in both x and y directions. The correct equations for momentum conservation are m1V = m1V1√2/2 + 2mV2√2/2 for the x direction and 0 = √2/2(m1V1 + 2mV2) for the y direction. The participant successfully simplified the problem and achieved a 90% score on their assignment.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with 2D vector components
  • Knowledge of trigonometric functions, specifically sin45° and cos45°
  • Basic algebra for solving equations
NEXT STEPS
  • Study advanced collision physics, focusing on elastic and inelastic collisions
  • Learn about momentum conservation in multi-body systems
  • Explore vector decomposition in physics problems
  • Practice solving similar 2D collision problems using different mass ratios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of 2D collision problems.

Visionary_
Messages
2
Reaction score
0

Homework Statement


We are only given variables in this problem, that is:
m1 travels at V, collides with 2m which is at rest
results in m going off at 45° at V1
2m also goes off at 45° down at V2
Calculate V1 and V2 in terms of V


Homework Equations


Ʃρx-initial=Ʃρx-final
Ʃρy-initial=Ʃρy-final
Use Linear combinations; sin45= √2/2=cos45

The Attempt at a Solution


x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)
 
Physics news on Phys.org
Visionary_ said:

Homework Statement


We are only given variables in this problem, that is:
m1 travels at V, collides with 2m which is at rest
results in m going off at 45° at V1
2m also goes off at 45° down at V2
Calculate V1 and V2 in terms of V


Homework Equations


Ʃρx-initial=Ʃρx-final
Ʃρy-initial=Ʃρy-final
Use Linear combinations; sin45= √2/2=cos45

The Attempt at a Solution


x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)

The equations you have written out are not quite right.

You have:

x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)

but were perhaps trying for

x direction:
m1V=m1V1√2/2+m2V2√2/2
y direction
0=√2/2(m1V1+m2V2)

Though perhaps you really did mean masses m and 2m ?
 
yes the second mass was 2m as in twice the mass,
i simplified the problem further and got a 90% on the paper so its fine :)
 

Similar threads

Replies
10
Views
6K
Replies
10
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
Replies
4
Views
2K