A FIREWORKS ROCKET IS MOVING WITH A SPEED OF 45.0 M/S. THE ROCKET SUDDENLY BREAKS INTO TWO PIECES OF EQUAL MASS, WHICH FLY OFF WITH VELOCITIES V1 AND V2. WHAT IS THE MAGNITUDE OF V1?
(THE ANGLE OF V1 WITH RESPECT TO THE ORIGINAL V DIRECTION IS 30 DEGREES WHILE THE ANGLE OF V2 IS 60 DEGREES.)
how did u get tis answer lisa,,
the answer is v1 = 2*45*cos(30) = 77.9 m/s
set x to initial rocket direction. so Vx=45m/s
1st conservation of linear momentum theorem in x direction:
m*45 = 1/2m*V1*cos30 + 1/2m*V2*cos60
cancel m and multiply by 2
90 = V1*cos30 + V2*cos 60
2nd conservation of linear momentum theorem in y direction. remember initial rocket Vy = 0
0 = 1/2*m*V1*sin30 - 1/2 *m*V2*sin60
put on opposite sides. 1/2 and m's cancel out.
therefore:
V1*sin30 = V2*sin60
Solve for V1 or V2 and plug in.
eventually if you solve for V1 and plug in:
90 = 1.5V2 + .5V2
2V2 = 90
V2 = 45m/s
Why did the fireworks rocket break into two pieces? It must have been explosive! Anyway, let's calculate the magnitude of V1.
To find the magnitude of V1, we need to use the law of conservation of momentum. According to this law, the total momentum before the explosion is equal to the total momentum after the explosion.
Since the rocket breaks into two equal mass pieces, each piece will have half of the original momentum.
The original momentum (p1) is given by mass (m1) times velocity (v):
p1 = m1 * v
After the explosion, the momentum of the first piece (p1) is given by:
p1 = (m1/2) * V1
Now, we can equate the two momenta:
m1 * v = (m1/2) * V1
Simplifying the equation, we get:
v = V1/2
Rearranging the equation, we find:
V1 = 2 * v
Plugging in the value of v (45.0 m/s), we get:
V1 = 2 * 45.0 m/s
So the magnitude of V1 is 90.0 m/s. I hope this answer didn't explode your mind!
To find the magnitude of velocity (V1) of one of the pieces of the broken fireworks rocket, we can use the law of conservation of momentum.
The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the rocket breaking into two pieces is a closed system.
The equation for conservation of momentum is:
m1 * V1 + m2 * V2 = m * V
Where:
m1 and m2 are the masses of the two pieces of the rocket (assumed to be equal).
V1 and V2 are the velocities of the two pieces after the explosion.
m is the total mass of the original rocket (before it broke into two pieces).
V is the original velocity of the rocket.
In this case, we are given:
V = 45.0 m/s (original velocity of the rocket)
m1 = m2 = m/2 (equal masses of the two pieces)
Now substitute the given values into the equation:
(m/2) * V1 + (m/2) * V2 = m * V
Since m1 = m2 = m/2, we can simplify the equation:
V1 + V2 = 2V
Since we know the angles of V1 and V2 with respect to the original V direction, we can use trigonometry to relate their magnitudes.
Let's consider V1 first. The given angle is 30 degrees. The magnitude of V1 can be found using the formula:
|V1| = V1_cos(theta)
Where theta is the angle between V1 and the original V direction.
So, |V1| = V1_cos(30 degrees)
Similarly, for V2:
|V2| = V2_cos(theta)
Where theta is the angle between V2 and the original V direction. In this case, theta is 60 degrees.
Since we already have V1 + V2 = 2V, we can substitute the magnitudes:
|V1| + |V2| = 2V
V1_cos(30 degrees) + V2_cos(60 degrees) = V
Now, we need to solve for V1.
From the given information, we know the rocket breaks into two equal masses. Since V1 and V2 are the velocities of the broken pieces, they should also be equal in magnitude:
|V1| = |V2|
Therefore, we can rewrite the equation as:
2|V1| = V
Simplifying further:
|V1| = V/2
Substituting the given value of V:
|V1| = 45.0 m/s / 2
Therefore, the magnitude of V1 is:
|V1| = 22.5 m/s