Collision of particles; centre of mass problem

[charmedbeauty]

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

y_x=h - 1/2 g(h/v_o))²

ie replacing t with (h/v_o)

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

I had

before p= m₁v₁+m₂v₂

= m₁(v_o-gt)+m₂(-gt)

and after p= (m₂m₂)v_x

deriving v_x from part e)

v_x= -g(h/v_o)

so after p= (m₁+m₂)(-g×h/v_o)

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

Thanks.

 

[Infinitum]

deriving v_x from part e)

v_x= -g(h/v_o)

so after p= (m₁+m₂)(-g×h/v_o)

But this v_x 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??

 

[charmedbeauty]

But this v_x 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.

 

[Infinitum]

ohh right thanks. infinitum. momentum.

Yep, that should give you the answer!

 

[asteriatic]

Your expression for the total momentum before the collision is correct. However, your expression for the total momentum after the collision is incorrect. In an inelastic collision, the total momentum is conserved, meaning that the total momentum before the collision is equal to the total momentum after the collision. Therefore, the expression for the total momentum after the collision should be the same as the expression for the total momentum before the collision, which is m1v1 + m2v2.

In order to solve for the velocities after the collision, we can use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. This can be expressed as:

m1v1 + m2v2 = (m1 + m2)vf

where vf is the final velocity of the particles after the collision. Since the collision is completely inelastic, the two particles will stick together and move with the same final velocity. We can then solve for vf:

vf = (m1v1 + m2v2)/(m1 + m2)

Substituting in the values for v1 and v2 from part (e), we get:

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

= [(m1vo + m2vo) - (m1 + m2)gt]/(m1 + m2)

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

= vo - gt

Therefore, the total momentum after the collision is:

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

This is the same expression as the total momentum before the collision, as expected.