Collision of particles; centre of mass problem

AI Thread Summary
The discussion focuses on a physics problem involving two particles: one fired upwards and another released from rest, leading to a completely inelastic collision. The user correctly derived the position of the particles after the collision but struggled with calculating the total momentum before and after the collision. They initially used the wrong velocity for the combined masses post-collision. The key takeaway is that momentum is conserved in completely inelastic collisions, which should guide the correct calculation of total momentum. Understanding this principle is essential for solving part f) of the problem accurately.
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Homework Statement



One particle, mass m1, is fired vertically upwards with an initial velocity v0 from the ground (height y = 0). At the same time (t = 0), another particle, mass m2, is released from rest at a height y = h, directly above the first, where h > 0. The particles are then in free fall near the Earth's surface. Air resistance and the rotation of the Earth may be neglected.

e) Assume that the particles collide and that the collision is very brief and completely inelastic. Write an expression for the position of the particles after the collision but before they hit the ground.

f) From the results above, derive an expression for the total momentum of the two particles both before and after the collision, but before they hit the ground.

Homework Equations





The Attempt at a Solution



for e) I had

yx=h - 1/2 g(h/vo))2

ie replacing t with (h/vo)

the expression is correct, but I am having trouble with part f).

I had

before p= m1v1+m2v2

= m1(vo-gt)+m2(-gt)

and after p= (m2m2)vx

deriving vx from part e)

vx= -g(h/vo)

so after p= (m1+m2)(-g×h/vo)

but this is wrong for part f), can someone please help with why?

Thanks.
 
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charmedbeauty said:
deriving vx from part e)

vx= -g(h/vo)

so after p= (m1+m2)(-g×h/vo)

But this vx is the velocity of the particle that was dropped just before the collision, and not the total velocity of both the masses together.

It is a completely inelastic collision, what quantity is conserved in such a collision??
 
Infinitum said:
But this vx is the velocity of the particle that was dropped just before the collision, and not the total velocity of both the masses together.

It is a completely inelastic collision, what quantity is conserved in such a collision??

ohh right thanks. infinitum. momentum.
 
charmedbeauty said:
ohh right thanks. infinitum. momentum.

Yep, that should give you the answer! :approve:
 
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