Question

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. Using the data shown in the drawing, find the masses of pieces 1 and 2.

m2 = 25.0° 11 m2 45.0° --- m3 = 1.30 kg 3.07 m/s

m1 =	kg
m2 =	kg

Answer 1

The condition for the total momentum parallel to the floor to be zero is:

$m_{1}\vec{v}_{1}+m_{2}\vec{v}_{2}=m_{3}\vec{v}_{3}$

We can write this by components:

$m_{1}v_{1}\sin (25^{\circ})=m_{2}v_{2}\cos (45^{\circ})$

m101 COS(25°) + m202 sin(45°) = m3u3

Replacing with the known values:

$m_{1}\times 2.8m/s\times \sin (25^{\circ})=m_{2}\times 1.75m/s\times \cos (45^{\circ})$

$1.183m_{1}=1.237m_{2}$

And

$m_{1}\times 2.8m/s\times \cos (25^{\circ})+m_{2}\times 1.75m/s\times \sin (45^{\circ})=1.3kg\times 3.07m/s$

$2.538m_{1}+1.237m_{2}=3.991$

Since from the first equation 1.237m₂ = 1.183m₁

$2.538m_{1}+1.183m_{1}=3.991$

$3.721m_{1}=3.991$

$m_{1}=1.073kg$

Finally if we go backwards to the first equation and solve for m₂ we get:

$m_{2}=1.026kg$