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.)
3 answers
the answer is v1 = 2*45*cos(30) = 77.9 m/s
how did u get tis answer lisa,,
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
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
0 answers