An object of mass 3M, moving in the +x direction at speed V_0, breaks onto two p

Question

An object of mass 3M, moving in the +x direction at speed V_0, breaks onto two pieces of mass M and 2m as shown in the figure below. Theta_1=40.0 degree and theta_2=25.0 degree, determine the final velocities of the resulting pieces in terms of V_0. Determine the x component of the velocity of the smaller piece V_1x in terms of V_O. (Express your answer to three significant figures.) Determine the y component of the velocity of the smaller piece V_1y in terms of V_O. (Express your answer to the three significant figures.) Determine the x component of the velocity of the larger piece V_2x in terms of V_O. (Express your answer of the three significant figures.) Determine the y component of the velocity of the larger piece V_2y in terms of V_O. (Express your answer of the three significant figures.)

Explanation / Answer

Here

theta1 = 40 degree

theta2 = 25 degree

Balancing the momentum in horizontal direction

2 M * v2 * cos(theta2) + M * v1 * cos(theta1) = 3 M* vo

2 * v2 * cos(25) + v1 * cos(40) = 3 * v0 ----(1)

in the vertical directio

2 M * v2 * sin(theta2) - M * v1 * sin(theta1) = 0

2 * v2 * sin(25) - v1 * sin(40) = 0 ---(2)

solving 1 and 2

v1 = 1.40 * v0

v2 = 1.06 * v0

1)

x component of smaller piece = v1 * cos(40)

x component of smaller piece = 1.40 * v0 * cos(40)

x component of smaller piece = 1.072 * v0

2)

y component of smaller piece = v1 * sin(40)

y component of smaller piece = 1.40 * v0 * sin(40)

y component of smaller piece = 0.90 * v0

3)

x component of larger piece = v2 * sin(theta2)

x component of larger piece = 1.06 * vo * sin(25)

x component of larger piece = 0.45 * v0

4)

y component of larger piece = 1.06 * v0 * cos(25)

y component of larger piece = 0.961 * v0

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