Question

A vase falls off a table and shatters into three fragments which move away horizontally across the floor. Fragment 1 has half the mass of the original vase and travels with speed 34.5 cm/s. Fragment 2 has one-quarter the mass of the original vase and travels with speed 26.6 cm/s in a direction perpendicular to Fragment 1.

What is the speed (in cm/s) of the third fragment?

Answer 1

here you have to apply conservation of momentum in x and y direction

mass of 3rd fragment = 1/4*m........here m is mass of original vase

Given , Fragment 1 has half the mass of the original vase and travels with speed 34.5 cm/s. Fragment 2 has one-quarter the mass of the original vase and travels with speed 26.6 cm/s in a direction perpendicular to Fragment 1.

therefore fragment 3 must move in such way that it negates balances momentum of both the fragments to 0.

in direction opposite to that of vase 1 , let speed of vase 3 be Vx

the

1/2*m*34.5 = 1/4*m*Vx

Vx = 69 cm/sec

In direction opposite to that of fragment 2 , let speed of vase 3 be Vy

then,

1/4*m*26.6 = 1/4*m*Vy

so Vy = 26.6 cm/sec

Now we know that fragment 1 and 2 moved perpendicular to each other ,the Vx and Vy also must be perpendicular to each other

so total speed of vase 3 will be

V = sqrt(Vx^2 + Vy^2) = 83.3580 cm/sec...........answer