How does momentum conservation depend on the collision type? You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the ball bounces off with no loss of speed, the clay sticks. What is the change in momentum of the clay and ball, respectively?

Respuesta :

Answer:

For ball P=- 2 x m x V.

For P= - m x V

Explanation:

As we know that

if external force is zero then linear momentum of system will be conserve.

[tex]P_i=P_f[/tex]

Rubber ball:

Given that speed of ball is V before the collision with ball.

The speed of the ball is remain after the collision with ball but the direction will reverse.

Lets take first ball is moving in positive x direction and after collision with ball it moves in negative x direction.

Initial momentum

[tex]P_i=m\times V[/tex]

Finall momentum

[tex]P_f=-m\times V[/tex]

So the change in the momentum

[tex]P=P_f-P_i[/tex]

P=- 2 x m x V.

For clay:

Initial velocity of clay is V in positive x direction and final velocity is zero because it stick with the wall

[tex]P_i=m\times V[/tex]

[tex]P_f=-m\times 0[/tex]

[tex]P_f=0[/tex]

So the change in the momentum

[tex]P=P_f-P_i[/tex]

P= - m x V.

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