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.
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?
how did u find
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.
To find the direction of the third piece, we need to determine the angle it makes with the horizontal.
Given the information provided, we know that one piece travels backward at an angle of 32.0° above the horizontal, and the second piece travels at an angle of 28.0° below the horizontal. Since the third piece flies backward as well, it will have an angle between these two angles.
Let's assume the angle made by the third piece with the horizontal is θ. Since the sum of angles in a triangle is 180°, we can write:
θ + 32.0° + 28.0° = 180°
θ + 60.0° = 180°
θ = 180° - 60.0°
θ = 120°
Therefore, the direction of the third piece is 120° above the horizontal.
To determine the direction of the third piece, we need to consider the velocities of the other two pieces and use the principle of conservation of momentum.
The total momentum before the collision is equal to the total momentum after the collision. Since there are no external forces acting on the system in the horizontal direction, the horizontal component of the momentum is conserved. Similarly, since there are no external forces acting on the system in the vertical direction, the vertical component of the momentum is conserved as well.
Let's denote the horizontal and vertical velocities of the third piece as Vx and Vy, respectively. We can then write the momentum conservation equations as follows:
Total initial horizontal momentum = Total final horizontal momentum
Total initial vertical momentum = Total final vertical momentum
For the horizontal component, we can write:
(m1 * V1x) + (m2 * V2x) = m3 * Vx
For the vertical component, we can write:
(m1 * V1y) + (m2 * V2y) = m3 * Vy
Since we know the masses and velocities of the first two pieces, we can substitute these values into the equations. Let's plug in the given values:
(2.8 kg * 10.1 m/s * cos(32°)) + (1.8 kg * 8.3 m/s * cos(28°)) = 6.8 kg * Vx
(2.8 kg * 10.1 m/s * sin(32°)) - (1.8 kg * 8.3 m/s * sin(28°)) = 6.8 kg * Vy
Now, we can solve these two equations simultaneously to find the values of Vx and Vy. The resulting values will give us the direction of the third piece.