A 6-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22 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 2430 N for 0.1 s. One piece of mass 2 kg travels backward at a velocity of 15 m/s and an angle of 32° above the horizontal. A second piece of mass 1 kg travels at a velocity of 7 m/s and an angle of 28° below the horizontal. What is the velocity of the third piece? (Let up be the +y positive direction and to the right be the +x direction.)

Initial momentum = Final momentum + impulse delvered by the wall.

You have all the information you need to solve for the momentum of the third pice.

Assume the force provided at the wall is in the -x direction

Muhammad

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Solution for q

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To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision.

1. Start by determining the initial momentum of the clay ball before it hits the wall. The initial momentum (p_initial) is given by the mass (m) multiplied by the initial velocity (v_initial).

p_initial = m * v_initial
p_initial = 6 kg * 22 m/s
p_initial = 132 kg·m/s

2. The ball shatters into three pieces, so we need to find the individual momenta of each piece after the collision.

Let the momentum of piece 1 (p1) be in the +x direction.
Let the momentum of piece 2 (p2) be in the -x direction.
Let the momentum of piece 3 (p3) be in the -y direction.

We can write the conservation of momentum equation as:

p_initial = p1 + p2 + p3

3. Now, let's find the values of p1 and p2 using the given information.

For piece 1: m1 = 2 kg and v1 = 15 m/s at an angle of 32° above the horizontal.

The x-component of momentum for piece 1 is given by:
p1x = m1 * v1x, where v1x = v1 * cos(theta1)

The y-component of momentum for piece 1 is given by:
p1y = m1 * v1y, where v1y = v1 * sin(theta1)

Substituting the values, we get:
p1x = 2 kg * 15 m/s * cos(32°)
p1x = 25.594 kg·m/s

p1y = 2 kg * 15 m/s * sin(32°)
p1y = 8.522 kg·m/s

Therefore, p1 = sqrt(p1x^2 + p1y^2)
p1 = sqrt((25.594 kg·m/s)^2 + (8.522 kg·m/s)^2)
p1 = sqrt(673.36 + 72.57)
p1 = sqrt(745.93)
p1 ≈ 27.30 kg·m/s

For piece 2: m2 = 1 kg and v2 = 7 m/s at an angle of 28° below the horizontal.

The x-component of momentum for piece 2 is given by:
p2x = m2 * v2x, where v2x = v2 * cos(theta2)

The y-component of momentum for piece 2 is given by:
p2y = m2 * v2y, where v2y = v2 * sin(theta2)

Substituting the values, we get:
p2x = 1 kg * 7 m/s * cos(28°)
p2x = 5.725 kg·m/s

p2y = 1 kg * 7 m/s * sin(28°)
p2y = 3.254 kg·m/s

Therefore, p2 = sqrt(p2x^2 + p2y^2)
p2 = sqrt((5.725 kg·m/s)^2 + (3.254 kg·m/s)^2)
p2 = sqrt(32.832 + 10.615)
p2 = sqrt(43.447)
p2 ≈ 6.59 kg·m/s

4. Now, let's substitute these values back into the conservation of momentum equation to find p3.

p_initial = p1 + p2 + p3
132 kg·m/s = 27.30 kg·m/s + 6.59 kg·m/s + p3

Rearranging the equation, we get:
p3 = 132 kg·m/s - (27.30 kg·m/s + 6.59 kg·m/s)
p3 = 98.11 kg·m/s

5. Finally, we need to find the velocity of piece 3.

The magnitude of the velocity of piece 3 (v3) can be found using the formula:
v3 = sqrt((p3x/m3)^2 + (p3y/m3)^2), where m3 is the mass of piece 3.

Since piece 3 travels only in the negative y-direction, we have:
p3x = 0
p3y = -m3 * v3y, where v3y is the y-component of the velocity of piece 3.

Substituting the values, we get:
v3 = sqrt((0/m3)^2 + (-98.11 kg·m/s / m3)^2)
v3 = sqrt(0 + (-98.11 / m3)^2)
v3 = 98.11 / m3

The mass of piece 3 is not given, so we cannot find the exact velocity of piece 3 without that information.