An object of mass 180 g initially at rest explodes into three pieces. One piece has mass 70 g and moves directly west at speed 14 m/s, and another piece has mass 50 g and moves directly south at speed 18 m/s. Find the speed and direction of the third piece.

the speed is 22.17m/s, and the direction is 42.56° from north to east.

Explanation:

The mass of the object is 180g, and it initialy has a velocity of 0m/s, so the total momentum is P = 0

In this problem we must use the conservation of the momentum, this means that the final momentum must be equal than the initial momentum.

we have 3 pieces:

One has a mass of m1 = 70g and v1 = 14m/s west.

other has a mass of m2 = 50g and v2 = 18m/s south.

this two pieces have a momentum of:

p1 = 70*14 (g*m/s) west = 980 (g*m/s) west

p2 = 50*18 (g*m/s) south = 900 (g*m/s) south.

now, the total momentum must be zero, so P3 = 980  (g*m/s)  east + 900 ( (g*m/s) north.

first, knowing that the total mass is 180g, we have that the mass of the third piece is:

m3 = 180g - 70g - 50g = 60g

now, for the momentum we will have that one component goes north and other east, if we define the angle A as the angle that goes from north to east (such that A = 0° means that we are moving directly to north) we will have:

p3 = 60g*v*(cos(A)) to north + 60g*v*sin(A)

then we have a system of equations:

60*v*cos(A) = 980

60*v*sin(A) = 900

Here we can use the quotient of equation 2 and equation 1, in this way we remove the variable v and we can solve it for A:

(60*v*sin(A) )/(60*v*cos(A)) = 900/980

tan(A) = 900/980

Atan(tan(A)) = A = Atan(900/980) = 42.56°

now with this angle we can find the value of v:

60*v*cos(42.56°) = 980

v = 980/(60*cos(42.56°) = 22.17 m/s

then the speed is 22.17m/s, and the direction is 42.56° from north to east.