Conservation of Momentum in Inelastic Collisions

[sauri]

Can anyone help me with this?

A steel wrecking ball of mass 200kg is fastned to a 10m long chain which is fixed at the far end to a crane. The ball is released when the chain is horizontal. At the bottom of the path the ball strikes a 1000kg car initally at rest on a frictionless surface. The collision is completely inelastic so that the ball and car move together just after the collision. What is the speed of the ball and car just after the collision.

 

[mezarashi]

I would be glad to, but according to the rules you need to demonstrate some effort in attempting to solve the problem first.

 

[sauri]

sure, but I have this notion that the mgh before collision should be equall to the mv^2/2 after collision. because it is inelatic.
So mgh=mv^2/2
200*9.8*10=1000*v^2/2
and we solve for v..
is this the correct way?

 

[phucnv87]

Using the law of conservation of momentum[/color]

 

[sauri]

Could you explain that a bit futher?. How can I use the conservation of momentum?. is it not the use of the GPE=1/2mv^2?

 

[HallsofIvy]

You can calculate the potential energy of the ball at the start and so find the kinetic energy, speed, and momentum of the ball just before the collision. Set momentum of car and ball together after the collision equal to that.

 

[sauri]

So then mgh=1/2mv^2 and that will equal mv. (200*9.8*10=1/2(200)*v^2). So we can find v just before the collison. Then the kinetic energy=momentum(mv) before collision. This will = momentum after collision. i.e mv of car + mv of ball. [(1000*v)+(200*v)]. Is this correct?

 

[Hootenanny]

So then mgh=1/2mv^2 and that will equal mv. (200*9.8*10=1/2(200)*v^2). So we can find v just before the collison. Then the kinetic energy=momentum(mv) before collision. This will = momentum after collision. i.e mv of car + mv of ball. [(1000*v)+(200*v)]. Is this correct?

NO! Kinetic energy does NOT equal momentum , they are entirely diffderent! You need to work out the velocity of the ball just before collison using:
$$v = \sqrt{2gh}$$
So the momentum of the recking ball is given by;
$$P = m\sqrt{2gh}$$.

The total amount of momentum must remain constant. Can you take it from there?