An object located at the origin and having mass M explodes into three pieces having masses M/4, M/3, and 5M/12. The pieces scatter on a horizontal frictionless xy-plane. The piece with mass M/4 files away with velocity 5.0 m/s at
37∘
above the x-axis. The piece with mass M/3 has velocity 4.0 m/s directed at an angle of
45∘
above the -x-axis. (a) What are the velocity components of the third piece? (b) Describe the motion of the CM of the system after the explosion.
Too complicated to do here, but here is the method.
Use conservation of momentum, in x and y directions.
Because the initial momenum, in x and y, was zero...
break each piece into x and y components of momentum.
add the momentumX and set to zero, same as momentumY set to zero.
you have two equations, two unknowns (velocities). Solve
Ok...I got that the angle theta from the positive x-axis is = 268 degrees (for the third piece).
Am I on the right track? Thank you!
55
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of angular momentum.
(a) To find the velocity components of the third piece, we need to sum up the momentum in the x and y directions separately for all the pieces before and after the explosion.
Before the explosion, the total momentum in the x-direction, P_x, is given by:
P_x = (M/4)(0 m/s) + (M/3)(4.0 m/s * cos(45°)) + (5M/12)(V_x)
Where V_x is the x-component of the velocity of the third piece.
Before the explosion, the total momentum in the y-direction, P_y, is given by:
P_y = (M/4)(5.0 m/s * sin(37°)) + (M/3)(4.0 m/s * sin(45°)) + (5M/12)(V_y)
Where V_y is the y-component of the velocity of the third piece.
Since the explosion happens at the origin, the total momentum before the explosion is zero, so we have:
0 = P_x + P_y
Now we can solve the above equations simultaneously to find the values of V_x and V_y.
(b) To describe the motion of the center of mass (CM) of the system after the explosion, we need to consider the momentum of the entire system before and after the explosion.
The CM of the system is given by:
CM_x = (M/4)(0 m) + (M/3)(0 m) + (5M/12)(V_x)
CM_y = (M/4)(0 m) + (M/3)(0 m) + (5M/12)(V_y)
Since the explosion happens at the origin, the CM of the system before the explosion is at the origin. Therefore, the CM after the explosion will continue to be at the origin.
The motion of the CM of the system remains unchanged after the explosion as there is no external force acting on the system.
Note: The angles given in the problem statement are in degrees, but trigonometric functions in mathematical calculations require angles in radians. Convert the given angles to radians before performing calculations.