Question

Question #9ebc9

Answer 1

No!

Explanation:

The law of conservation of momentum says that the overall change in momentum is zero.

`DeltavecP=0`

where `P` is the total linear momentum

`=>vecP_f-vecP_i=vec0`

Or equivalently, we might say that the momentum before is equal to the momentum after.

`=>vecP_f=vecP_i`

We know that momentum obeys the principle of superposition, so:

`vecP=vecp_"total"=sumvecp=vecp_1+vecp_2+...+vecp_n`

Finally, we know that `p=m*v`, where `m` is the mass of the object and `v` is the object's velocity.

Assuming the ball begins at rest, it has an initial momentum of zero as `v_i=0`.

When the ball is thrown upward, it gains a momentum of `mv_i`.

However, as it climbs higher, it loses velocity due to the negative acceleration imposed upon it by gravity `(veca=-g)`, and therefore its momentum decreases proportionately. Eventually it reaches a maximum altitude where `v=0`, and consequently `p=0`.

So, the ball began from rest at `p=0` and ended at rest with `p=0`, if we only look at the first half of the motion. We can see that conservation of momentum is not violated.

If we continued to analyze the situation as the ball falls back down, we know it begins with momentum `p_i=0` at its maximum altitude, gains momentum as its velocity increases (with motion now in the same direction as the acceleration), and then hits the ground and comes to rest with `v_f=0` and therefore `p_f=0`.

In both cases, `DeltaP=0`.

Answer 2

No, the principle of conservation or momentum applies only to isolated systems.

Explanation:

The fine print of the principle states that the system must be free from "outside forces". Ref:
http://www.physicsclassroom.com/class/momentum/Lesson-2/Isolated-Systems

If you do not include the Earth as part of the system, that exempts the thrown ball from application of the principle of conservation or momentum. If you include the Earth as part of the "system", there is no change in total momentum.

I hope this helps,
Steve