Using the law of conservation of kinetic energy

Click For Summary
SUMMARY

The discussion centers on solving a physics problem involving two objects, A and B, with masses m and 3m, respectively, moving towards each other with velocities v and -v. The user initially misapplies the law of conservation of kinetic energy and momentum, leading to incorrect conclusions about the velocities after impact. The correct approach involves factoring the equation rather than dividing by (v2), which clarifies that the second object's velocity (v2) can be determined accurately. The final solutions are v2 = 0 and v1 = -2v, indicating a misunderstanding in the initial calculations.

PREREQUISITES
  • Understanding of the law of conservation of momentum
  • Knowledge of the law of conservation of kinetic energy
  • Familiarity with algebraic manipulation of equations
  • Basic principles of classical mechanics
NEXT STEPS
  • Review the law of conservation of momentum in elastic collisions
  • Study the law of conservation of kinetic energy in inelastic collisions
  • Practice solving collision problems in one dimension
  • Explore the implications of mass and velocity on collision outcomes
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding collisions and energy conservation principles.

Moe_the_Genius
Messages
13
Reaction score
0
HI,
I am wondering how to solve this physics question:

-We have an object A and we have an object B
-We place these objects at a distance from each other.
-Object A has mass m.
-Object B has mass 3m.
-Object A moves toward object B with velocity v.
-Object B moves toward object A with velocity -v.

Q; What is the velocity of each object after impact?
--------------------------------------------------------
Can you tell me what I am doing wrong here: (I don't know where I have made a mistake)

(v1) = velocity of Object A after impact.
(v2) = velocity of Object B after impact.

Total momentum = mv-3mv= -2mv = m(v1) + 3m(v2)

Dividing both sides by m we find that:
-2v = (v1) + 3(v2)

and solving for v1 we find,

(v1) = -2v-3(v2) (1)

Using the law of conservation of kinetic energy we also get:

(.5)mv^2 + (.5)(3m)v^2 = (.5)m(v1)^2 + (.5)(3m)(v2)^2

Now divide by .5 and then by m to come up with:

4v^2 = (v1)^2 + 3(v2)^2 (2)

Inserting v1 from (1) into (2):

4v^2 = (-2v-3(v2))^2 + 3(v2)^2
4v^2 = 4v^2 + 12v(v2) + 9(v2)^2 + 3(v2)^2
-12v(v2) = 12 (v2)^2

and finally (dividing by (v2) and 12),
(v2)=-v

plugging (v2)=-v into (2) to come up with:
(v1)= v

It doesn't even make sense or look right and I have been trying to determine where I have made a mistake for days! :confused: :confused:
Thanks for any help...
 
Physics news on Phys.org
You only have one of the solutions, namely the one where they fail to interact. The other solution is v2=0, v1=-2v.
 
Wow, you really caught my mistake!
I understand now, instead of dividing by (v2) I should have brought -12v(v2) to the other side and factored to find (v2). Thanks a bunch.
 

Similar threads

Replies
5
Views
3K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
10
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
40
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K