Answer:
Immediately after throwing the backpack away, twin A would be moving away from twin B at approximately .
Initially, twin B would not immediately be moving. However, after the backpack hits her, she would move away from twin A at approximately if she held onto the backpack.
Explanation:
Consider this scenario in three steps:
- Step one: twin A is carrying the backpack.
- Step two: twin A throws the backpack away; the backpack is en route to twin B;
- Step three: twin B starts to move after the backpack hits her.
Since all external forces are ignored, momentum should be conserved when changing from step one to step two, and from step two to step three.
<h3>From step one to step two</h3>In step one, neither twin A nor the backpack is moving. Their initial momentum would be zero. That is:
.
.
Therefore:
.
In step two, the backpack is moving towards twin B at . Since the mass of the backpack is
, its momentum at that point would be:
.
Momentum is conserved when twin A throws the backpack away. Hence:
.
Therefore:
.
The mass of twin A (without the backpack) is . Therefore, her velocity in step two would be:
.
Note that while the velocity of the backpack is assumed to be greater than zero, the velocity of twin A here is less than zero. Since the backpack is moving towards twin B, it can be concluded that twin A is moving in the opposite direction away from twin B.
<h3>From step two to step three</h3>In step two:
since twin B is not yet moving.
from previous calculations.
Assume that twin B holds onto the incoming backpack. Thus, the velocity of the backpack and twin B in step three will be the same. Let denote that velocity.
In step three, the sum of the momentum of twin B and the backpack would thus be:
.
Simplify to obtain:
.
Momentum is conserved when twin B receives the backpack. Therefore:
.
Therefore:
.
In other words, if twin B holds onto the backpack, then (after doing so) she would be moving away from twin A at approximately .