How Does the Second Fragment's Velocity Change After a 2D Explosion?

[satasonic]

Homework Statement

A 4.00-kg cannon ball is flying at 18.5 m/s
[0º] when it explodes into two fragments.
One 2.37-kg fragment (A) goes off at
19.7 m/s [325º]. What will be the velocity of
the second fragment (B) immediately after
the explosion? Assume that no mass is lost
during the explosion, and that the motion
of the fragments lies in the xy plane.

Homework Equations

p = mv

The Attempt at a Solution

K so i got my graph drawn, but because this is an explosion i have no idea how to set up the x and y equations, any help?

 

[rl.bhat]

You have to apply the conservation momentum. The explosion does not make any difference.
Resolve the momentum Pa and Pb along x and y axis. And proceed.

 

[tomcue]

I would approach this problem by first identifying the given parameters and variables. In this case, we have a cannon ball with a mass of 4.00 kg traveling at a velocity of 18.5 m/s in the x-direction before it explodes into two fragments. One fragment, A, has a mass of 2.37 kg and a velocity of 19.7 m/s at an angle of 325º. We are also told that no mass is lost during the explosion and that the motion of the fragments lies in the xy plane.

Next, I would use the conservation of momentum principle to solve for the velocity of the second fragment, B. This principle states that the total momentum before the explosion is equal to the total momentum after the explosion. In this case, we can set up the equation as follows:

(m1v1 + m2v2)before = (m1v1 + m2v2)after

Where m1 and v1 represent the mass and velocity of the cannon ball before the explosion, and m2 and v2 represent the mass and velocity of the second fragment, B, after the explosion.

We can also use the momentum vector notation to represent the velocities and angles of the fragments. The velocity of the cannon ball before the explosion can be written as (18.5 m/s)[0º], and the velocity of fragment A can be written as (19.7 m/s)[325º]. Using this notation, our equation becomes:

(4.00 kg)(18.5 m/s)[0º] + (2.37 kg)(0 m/s)[0º] = (4.00 kg)(vB)[θB] + (2.37 kg)(vB)[θB]

Solving for vB, the velocity of the second fragment, we get:

vB = (4.00 kg)(18.5 m/s)[0º] + (2.37 kg)(19.7 m/s)[325º] / (4.00 kg + 2.37 kg)

vB = (74.00 kg•m/s)[0º] + (46.689 kg•m/s)[325º] / 6.37 kg

vB = (74.00 kg•m/s + 46.689 kg•m/s) / 6.37 kg

vB = 15.943 m/s

Therefore, the