Question 2.

3 Consider a box that explodes into two pieces while moving with
Fig. 9-24
a constant positive velocity along an x axis. If one piece, with mass :
m1, ends up with positive velocity v1, then the second piece, with :
mass m2, could end up with (a) a positive velocity v2 (Fig. 9-25a), (b) :
a negative velocity v2 (Fig. 9-25b), or (c) zero velocity (Fig. 9-25c). Rank those three possible results for the second piece according to
:
the corresponding magnitude of v1, greatest first. v2v1v2 v1 v1

(a) (m₁+m₂)v=m₁v₁+m₂v₂

v₁(a)= {(m₁+m₂)v-m₂v₂}/m₁
(b) (m₁+m₂)v=m₁v₁-m₂v₂
v₁(b)= {(m₁+m₂)v+m₂v₂}/m₁
(c) (m₁+m₂)v=m₁v₁
v₁(c)= (m₁+m₂)v/m₁
v₁(b) > v₁(c)> v₁(a)

Well, I gotta say, it sounds like that box is having a real blast! But let's break it down and rank those possibilities.

First off, let's consider option (c) - zero velocity for the second piece. Now, if the first piece is zooming along with a high positive velocity, it would make sense for the second piece to come to a halt. Kinda like a "stop, drop, and roll" situation, but without the fire. So, the magnitude of v1 would be the greatest in this case.

Next up, we have option (b) - a negative velocity for the second piece. Now, if the first piece is moving forward, it's possible that the second piece is like, "nah, I'm going in reverse." So, the magnitude of v1 would still be greater in this case.

And finally, we have option (a) - a positive velocity for the second piece. Here's where things get interesting. If the first piece is traveling at a really high positive velocity, the second piece could potentially still go forward, but at a slower speed. Like a speed limit for shattered boxes. So, the magnitude of v1 would be the smallest in this scenario.

So, to rank them in order of magnitude of v1, it would be (c) zero velocity, (b) negative velocity, and (a) positive velocity. I hope that clears things up and brings a little explosion of laughter to your day!

To rank the three possible results for the second piece according to the magnitude of v1, let's analyze the situation step-by-step:

Step 1: Understand the scenario
We have a box moving with a constant positive velocity along the x-axis. The box explodes into two pieces, with masses m1 and m2. One piece, with mass m1, ends up with a positive velocity v1.

Step 2: Analyze the scenarios
(a) Positive velocity v2 (Fig. 9-25a): In this scenario, the second piece, with mass m2, also has a positive velocity v2. The magnitude of v1 is positive, and the magnitude of v2 can also be positive. This means that the magnitude of v1 is greater than the magnitude of v2. Thus, we can rank this scenario as the third in terms of the magnitude of v1.

(b) Negative velocity v2 (Fig. 9-25b): In this scenario, the second piece, with mass m2, has a negative velocity v2. The magnitude of v1 is positive, and the magnitude of v2 is negative. Since the magnitude of v1 is greater than the magnitude of v2, we can rank this scenario as the second in terms of the magnitude of v1.

(c) Zero velocity v2 (Fig. 9-25c): In this scenario, the second piece, with mass m2, has zero velocity. The magnitude of v1 is positive, but the magnitude of v2 is zero. Since the magnitude of v1 is greater than the magnitude of v2, we can rank this scenario as the first in terms of the magnitude of v1.

Ranking the scenarios from greatest to least in terms of the magnitude of v1:
1. Zero velocity v2 (Fig. 9-25c)
2. Negative velocity v2 (Fig. 9-25b)
3. Positive velocity v2 (Fig. 9-25a)

To rank the three possible results for the second piece according to the corresponding magnitude of v1, we need to analyze the conservation of momentum.

According to the law of conservation of momentum, the total momentum before and after an explosion remains constant, as long as no external forces act on the system. Mathematically, it can be written as:

m1 * v1(initial) + m2 * v2(initial) = m1 * v1(final) + m2 * v2(final) ........(1)

where,
m1 = mass of the first piece,
v1(initial) = initial velocity of the first piece,
m2 = mass of the second piece,
v2(initial) = initial velocity of the second piece,
v1(final) = final velocity of the first piece,
v2(final) = final velocity of the second piece.

We are given that the first piece ends up with a positive velocity v1. Now, let's analyze each possible scenario for the second piece:

a) Positive velocity v2:
In this case, both pieces move in the same direction with positive velocities. Therefore, the magnitudes of both v1 and v2 are positive. The magnitude of v1 is greater than v2.

b) Negative velocity v2:
In this case, the second piece moves in the opposite direction with a negative velocity. The magnitude of v1 will still be greater than the magnitude of v2 because it has a positive value.

c) Zero velocity v2:
In this case, the second piece comes to a stop (zero velocity). The magnitude of v1 will be greater than zero since it is positive.

Ranking the possibilities in terms of the corresponding magnitude of v1, from greatest to least, would be:

(a) Positive velocity v2 (v1 > v2 > 0)
(b) Zero velocity v2 (v1 > 0, v2 = 0)
(c) Negative velocity v2 (v1 > |v2|)

So, the ranking would be (a), (c), (b) in terms of the corresponding magnitudes of v1.

a>b>c