By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing find (a) the mass of piece 1 and (b) the mass of piece 2.
when the plate falls, its momentum is made only vertical component and component parallel to the floor are absent. The momentum parallel to the floor is zero, so only the momentum of the plate must be conserved in perpendicular directions.
Along the x direction :
-m1v1sinθ + m2v2cosθ = 0
Along the y direction:
m1v1cosθ + m2v2sinθ - m3v3 = 0
On substituting the values and solving the question,
-m1×3×sin25 + m2×1.79×cos45 = 0
m1×3×cos25 + m2×1.79×sin45 - 1.30×3.07 = 0
-1.268m1 + 1.266m2 = 0
2.719m1 + 1.266m2 = 3.991
Thus m1 = 1.00kg and m2 = 1.00kg
By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart...
By accident, a large plate is dropped and breaks into three
pieces. The pieces fly apart parallel to the floor, with
v1 = 2.80 m/s and v2 = 1.75
m/s. As the plate falls, its momentum has only a vertical
component, and no component parallel to the floor. After the
collision, the component of the total momentum parallel to the
floor must remain zero, since the net external force acting on the
plate has no component parallel to the floor....
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