Elastic collision, frictionless surface

AI Thread Summary
In an elastic collision on a frictionless surface, two air hockey pucks collide: one with a mass of 0.050 kg moving right at 5.0 m/s and the other with a mass of 0.10 kg moving left at 2.0 m/s. The conservation of momentum and conservation of energy equations are applied to find the final velocities after the collision. The first step involves setting up the momentum equation and solving for one of the final velocities in terms of the other. Substituting this expression back into the momentum equation allows for solving the unknowns. The discussion emphasizes deriving the equations correctly and plugging in values to find the final velocities.
synergix
Messages
177
Reaction score
0

Homework Statement


Two air hocky pucks collide head-on elastically on a frictionless surface. The smaller puck has a mass of 0.050 kg and is moving to the right at 5.0 m/s while the larger puck has a mass of 0.10 kg and is moving to the left with a speed of 2.0 m/s. Find the velocity of each after the collision.

Homework Equations


M1V1+M2V2=M1V'1+M2V'2
.5M1V1^2+.5M2V2^2=.5M1V'1^2+.5M2V'2^2

The Attempt at a Solution


I am pretty sure I must derive an equation from conservation of momentum and conservation of energy.
 
Physics news on Phys.org
That would be the way to do it.

Select a positive X direction and be careful of signs.

Two equations, 2 unknowns.
 
Cant anyone help me?
 
ok, could you give me an idea of what first step I should take?
 
first I take the conservation of momentum:
M1V1+M2V2=M1V'1+M2V'2

Then I solved for V'1

V'1=(M1V1+M2V2-M2V'2)/M1
Then I substituted this for V'1 in the original equation:
M1V1+M2V2=M1[(M1V1+M2V2-M2V'2)/M1]+M2V'2

so now the only unknown variable is V'2. I have little expierence deriving equations so I am not confident this is correct. though it certainly seems to be. to me anyways.
 
My first step would be to start plugging in the values and deriving the actual equations for this situation.

With the equations in hand then solve for the 2 velocities in the usual manner.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top