After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 5.33-kg ball is shot into the air with an initial speed of 11.3 m/s at a 42.9° angle; it explodes at a peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of 3.45 m/s. Another piece travels straight back with a speed of 2.39 m/s. What is the velocity of the third piece?
A = 3.45 m/s[90o]
B = 2.39 m/s[270o]
C = V m/s
3.45[90o]+2.39[270]+V = 0 @ the peak of
its' trajectory.
3.45*cos90+2.39*cos270+i3.45*sin90+i2.39*sin270 + V = 0
0+0 + 3.45i-2.39i + V = 0
0 + 1.06i + V = 0
V=-1.06i = 1.06 m/s[270o]=1.06m/s[-90o].
To solve this problem, we can apply the principles of projectile motion and conservation of momentum.
First, let's analyze the motion of the bowling ball before it explodes. We can break down its initial velocity into horizontal and vertical components.
The initial velocity of the ball, 11.3 m/s, can be split into the vertical component (V_y) and horizontal component (V_x).
V_y = 11.3 m/s * sin(42.9°)
V_y = 11.3 m/s * 0.682
V_y = 7.7116 m/s
V_x = 11.3 m/s * cos(42.9°)
V_x = 11.3 m/s * 0.731
V_x = 8.2643 m/s
We can now calculate the time it takes for the ball to reach its peak, which is the point of explosion.
We know that the vertical component of velocity decreases uniformly until the ball reaches its highest point where its vertical velocity becomes 0. At that point, the ball starts to fall down.
Using the equation V_fy = V_iy + a_y * t, where V_fy is the final vertical velocity (0 m/s), V_iy is the initial vertical velocity (7.7116 m/s), a_y is the acceleration due to gravity (-9.8 m/s^2), and t is the time, we can solve for t.
0 m/s = 7.7116 m/s - 9.8 m/s^2 * t
9.8 m/s^2 * t = 7.7116 m/s
t = 7.7116 m/s / 9.8 m/s^2
t = 0.7869 s
Now, let's calculate the vertical displacement (d_y) at the peak of the trajectory.
d_y = V_iy * t + 0.5 * a_y * t^2
d_y = 7.7116 m/s * 0.7869 s + 0.5 * (-9.8 m/s^2) * (0.7869 s)^2
d_y = 6.061 m
At the peak of the trajectory, the vertical displacement is 6.061 m.
Now, let's consider the horizontal motion. The horizontal displacement (d_x) does not change during the flight because there is no horizontal acceleration. Therefore, the horizontal displacement remains constant.
Now, let's analyze the explosion of the ball. We are given two pieces that move in the upward and backward directions.
One piece moves straight up with a speed of 3.45 m/s. Since the original ball had a vertical velocity of 7.7116 m/s at its peak, we can calculate the change in velocity (ΔV_y) for this upward piece.
ΔV_y = 3.45 m/s - 0 m/s
ΔV_y = 3.45 m/s
Similarly, the other piece moves straight back with a speed of 2.39 m/s. Since the original ball had a horizontal velocity of 8.2643 m/s, we can calculate the change in velocity (ΔV_x) for this backward piece.
ΔV_x = 8.2643 m/s - 0 m/s
ΔV_x = 8.2643 m/s
Now, let's use the principle of conservation of momentum to find the velocity of the third piece. We will assume the mass of each piece is m.
The initial momentum of the system is equal to the sum of the final momenta of the three pieces.
Initial momentum = Final momentum of upward piece + Final momentum of backward piece + Final momentum of third piece
Since the initial momentum of the system is zero (as there was no initial horizontal motion), the equation becomes:
Final momentum of upward piece + Final momentum of backward piece + Final momentum of third piece = 0
The final momentum of the upward piece is m * ΔV_y, and the final momentum of the backward piece is m * ΔV_x.
Therefore, m * ΔV_y + m * ΔV_x + Final momentum of third piece = 0
Since the mass of the third piece is also m, we can rewrite this equation as:
ΔV_y + ΔV_x + Final velocity of third piece = 0
Final velocity of third piece = -(ΔV_y + ΔV_x)
Final velocity of third piece = -(3.45 m/s + 8.2643 m/s)
Final velocity of third piece = -11.7143 m/s
Therefore, the velocity of the third piece is approximately -11.7 m/s. Note that the negative sign indicates that the third piece is moving in the opposite direction of the original horizontal motion.