How Does Conservation of Momentum Apply in a Three-Part Explosion?

[falcon0311]

I'm having trouble solving this:

Vessel at rest explodes. There's 3 pieces. Two pieces, one with twice the mass of the other, fly off perpendicular to one another with the same speed of 31.4 m/s. The third piece has three times the mass of the lightest piece. Find the magnitude and direction of its velocity immediately after the explosion. (Specify direction by giving the angle from the line of travel of the least massive piece.)

I realize the masses can be considered m, 2m, and 3m, and m can be substituted for a mass like 10kg. My initial sketch looks something like this:

-------- m ( 31.4 m/s )
|
|
|
|
2m ( 31.4 m/s )

P = p1 + p2 + p3 = m(31.4m/s) + 2m(31.4m/s) + 3m(v3)

The third momentum vector I realize is not just the bisected angle plus 180. My mind is having trouble coming up with that vector. Any help in solving the problem is much appreciated.

I'm getting 23.4 m/s (approx) at 117 degrees from m.

-Jacob

 

[HallsofIvy]

Trying to do vectors with trigonometry drives me up the wall!

What I would do is this: set up a coofdinate system so that the lightest piece moves along the positive x-axis. Since the second piece moves at right angles, we can take that to be the y-axis. Taking m to be the mass of the lightest piece, its momentum vector is
<34.1m, 0> and the momentum vector of the second piece is <0, (34.1)(2m)>= <0, 64.2m>. Their total momentum is <34.1m, 64.2m>. Since the total momentum of the system is 0, the momentum vector of the third piece is <-34.1m, -64.2m>. Since that third piece has mass 3m, its velocity vector is <-34.1/3, -64.2/3>= <-11.4,-22.7>.

The "length" of that last vector, i.e. the speed of the third piece, is &radic;((11.4)²+ (22.7)²)= 25.5 m/s. Its angle with the positive x-axis (i.e. relative to the motion of the first piece) is arctan(22.7/11.4)= 62.7 degrees except that it is in the third quadrant (both components are negative). The last piece moves at 62.7+ 90= 152.7 degrees measured clockwise from the motion of the first piece or
270-62.7= 207.3 degrees measured counterclockwise.

 

[M98Ranger]

Hi Jacob,

I understand that you are having trouble solving this problem involving momentum and explosions. Let's break it down step by step to help you find the solution.

First, we need to understand the concept of momentum. Momentum is the product of an object's mass and velocity and is a vector quantity. This means that it has both magnitude and direction.

In this scenario, we have a vessel at rest that explodes into three pieces. The first two pieces have masses of m and 2m and fly off perpendicularly to each other at a speed of 31.4 m/s. This means that their momentum vectors are equal in magnitude (m x 31.4 m/s) and opposite in direction, as they are flying off in opposite directions.

The third piece has a mass of 3m, and we need to find its magnitude and direction of velocity immediately after the explosion. To do this, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant.

So, we can set up an equation using the initial momentum of the system (before the explosion) and the final momentum of the system (after the explosion). This can be written as:

Pinitial = Pfinal

Since the vessel was at rest before the explosion, the initial momentum of the system is 0. This means that the final momentum of the system must also be 0.

Pfinal = m(31.4 m/s) + 2m(31.4 m/s) + 3m(v3) = 0

We can rearrange this equation to solve for v3, the velocity of the third piece:

v3 = -(m(31.4 m/s) + 2m(31.4 m/s)) / 3m

v3 = -20.93 m/s

The magnitude of the velocity of the third piece is 20.93 m/s, and its direction is in the opposite direction of the sum of the initial momenta of the first two pieces. This can be represented by a vector diagram, where the vectors representing the momenta of the first two pieces are added together and then the third piece's velocity vector is drawn in the opposite direction.

To find the angle of this vector, we can use the inverse tangent function:

θ = tan^-1(20.93/31.4) = 33.6 degrees

This means that the direction of the third