An m = 6.80-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22.0 m/s and shatters into three pieces, which all fly backward, as shown in the figure. The wall exerts a force on the ball of 2670 N for 0.110 s. One piece of mass m1 = 2.80 kg travels backward at a velocity of 10.1 m/s and an angle of = 32.0° above the horizontal. A second piece of mass m2 = 1.80 kg travels at a velocity of 8.30 m/s and an angle of 28.0° below the horizontal.

Question

I found the velocity of the third piece to be 48.738 m/s

but i need help finding:
What is the direction of the third piece?

Answer 1

how did u find

Answer 2

From conservation of momentum:

initial momentum of ball= (total momentum of the three balls)+(impulse of the wall)

First, lets find the momentum of the entire system in the x-axis.

(6.80)(22.0)cos(0)= 2670*(0.11)cos(0)+((2.80)(10.1)cos(32))+(1.80)(8.30)cos(-28)+(6.80-2.80-1.80)(v)cos(theta)

v*cos(theta)=-82.397289634823

Now lets take the total momentum in the system in the y-axis

(6.80)(22.0)sin(0)= (2670)(0.110)sin(0)+(2.80)(10.1)sin(32)+(1.80)(8.30)sin(-28)+(6.8-2.80-1.80)v*sin(theta)

v*sin(theta)=3.623732565679
v*cos(theta)=-82.397289634823
tan(theta)= (3.6237325665679/-82.397289634823)
theta= -2.5181760257216 degrees

In other words the third ball will go at the direction 2.5181760257216 degrees below the horizontal.