Angle of particles after elastic collision

Click For Summary
In an elastic collision between two particles of masses M1 and M2, where M1 is greater than M2, the angle of M1 after the collision is constrained by the relationship arcsin(M2/M1). The conservation of momentum and kinetic energy equations are essential in deriving the post-collision velocities of both particles. The velocities after the collision are given by v1` and v2`, which depend on the mass ratio and initial velocity. To determine the maximum angle for M1, projection methods can be applied to relate the angle to the velocity components. Understanding the critical mass difference where M1 exceeds M2 is key to solving the problem.
ajl1989
Messages
7
Reaction score
0

Homework Statement


A particle of mass M1 collides elastically with a particle of mass M2 at rest. Show that if M1>M2 then the angle of M1 after the collision cannot exceed the value arcsin(M2/M1).


Homework Equations


conservation of momentum: M1v1=M1v1`+M2v2`
conservation of kinetic energy: (1/2)M1v12=(1/2)M1v1`2+(1/2)M2v2`2


The Attempt at a Solution


I've figured out that the velocities after collision are:
v1`=v1*[(M1-M2)/(M1+M2)]
and
v2`=v1*[(2M1)/(M1+M2)]
but I don't know how to find the maximum angle of M1 after collision. Please help
 
Physics news on Phys.org
ajl1989 said:

Homework Statement


A particle of mass M1 collides elastically with a particle of mass M2 at rest. Show that if M1>M2 then the angle of M1 after the collision cannot exceed the value arcsin(M2/M1).

Homework Equations


conservation of momentum: M1v1=M1v1`+M2v2`
conservation of kinetic energy: (1/2)M1v12=(1/2)M1v1`2+(1/2)M2v2`2

The Attempt at a Solution


I've figured out that the velocities after collision are:
v1`=v1*[(M1-M2)/(M1+M2)]
and
v2`=v1*[(2M1)/(M1+M2)]
but I don't know how to find the maximum angle of M1 after collision. Please help

what's the critical point between when M1 > M2 is true, and when it's no longer true?

what will M1 - M2 be at that point?

to get angle into your equation, think projection.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K