A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. immediately after the impact, a 200-g piece moves along the + x-axis with a speed of 2.00 m/s, a 235-g piece moves along the + y-axis with a speed of 1.50 m/s. the third piece has a mass of 100 g. what is the speed of the third piece?

We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_ (f) = p_ (1) + p_ (2) + p_ (3) = 0.

So p_ (1x) = m1v1 = 0.2 * 2 = 0.4 Also p_ (2x) = m2v2 = 0 and p_ (3x) = m3v3 = 0.1 * v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed

Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_ (fy) = p_ (1y) + p_ (2y) + p_ (3y) = 0.

So p_ (1y) = 0, p_ (2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_ (3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.

So p_ (fx) = p_ (1x) + p_ (2x) + p_ (3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4

Also p_ (fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = - 3.525

So v_3x = - 4 and v_3y = 3.525.

The speed is their resultant = âš (-4) ^2 + (-3.525) ^2 = 5.335