Respuesta :
Answer:
The distance from the launch site is x [tex]= 6526.436m[/tex]
Explanation:
In this question we are provided with some details which include
The mass of the Cannonball M = 29.3 kg
The large piece of the ball [tex]M_A=\frac{2}{3} * 29.3 = 16.7333 Kg[/tex]
The mass of the small piece of the ball [tex]M_B= \frac{1}{2} * 29.3 = 8.3667 Kg[/tex]
From motion on a projectile, mathematically the time to reach the maximum height
[tex]\frac{u sin \theta }{g} = \frac{29.8}{2}[/tex] given that g =9.8 m/s
The 29.8 is divided by 2 because it is time taken for the journey of the cannonball
=> [tex]u sin (45) = 146.02[/tex]
[tex]u = 206.47 m/s[/tex]
The maximum height [tex]H = \frac{u^2 sin^2 \theta }{2g} = \frac{(206.47)^2* (sin 45)^2}{2*9.8 } = 1087.849 m[/tex]
Now after the explosion
Let [tex]M_A[/tex] have a velocity of [tex]V_A[/tex]
And [tex]M_B[/tex] have a velocity of [tex]V_B[/tex] [tex]= -ucos \theta i= -146.02i m/s[/tex]
We are able to calculate for the small piece of the cannonball because it took the path of the main cannonball
Considering the law of conservation of momentum
[tex]M_A V_A + M_B V_B = M(ucos \theta)[/tex]
[tex]\frac{2}{3} M * V_A + \frac{1}{3} M (146.02i) = M(146.02i)[/tex]
[tex]146.02i + 2M_A = 438.06i[/tex]
[tex]2V_A = 584.08i[/tex]
[tex]V_A = 292.04i \ m/s[/tex]
Mathematically the expression for the range
[tex]R = \frac{u^2 sin2 \theta }{g} = \frac{(206.470)^2 sin (2*45)}{9.8} = 4350.08m[/tex]
Now to obtain the distance of the larger part from the original source of projection
on landing
We make use of this projectile expression
[tex]x = \frac{R}{2} + v_A \sqrt{\frac{2H}{g} } = \frac{4350.08}{2} + (292.04)\sqrt{\frac{2(1087.849)}{9.8} } = 2175.04 + 4351.396[/tex]
[tex]= 6526.436m[/tex]
The motion of the cannonball before and after it explodes is a projectile
motion, that has a vertical velocity determined by gravity.
- The larger piece travels approximately 1,059.48 meters from the launch site
Reasons:
Mass of cannon ball, m = 29.3 kg
Angle of elevation at which it was launched = 45°
Point at which the cannon ball explodes = The top of its trajectory
Number of pieces formed following explosion = 2
Mass of smaller piece, m₂ = [tex]\displaystyle \mathbf{\frac{1}{3} \times m}[/tex]
Mass of larger piece, m₃ = [tex]\displaystyle \mathbf{\frac{2}{3} \times m}[/tex]
Point where the smaller piece lands = Point where cannonball was launched
Time after which the small piece lands = 12.0 s
Required:
The distance travelled from the launch site the larger piece travels.
Solution:
In the free fall motion of the cannonball, the time it takes the small piece to
reach maximum height, [tex]t_{max}[/tex] is the same as the time it takes the piece to
return to the ground level, therefore;
2 × [tex]t_{max}[/tex] = Time to reach maximum height + Time back to ground = 12 s
[tex]\displaystyle t_{max} = \frac{12.0 \, s}{2} = 6.0 \, s[/tex]
[tex]\displaystyle t_{max}= \mathbf{\frac{u \cdot sin(\theta)}{g}}[/tex]
Where;
g = The acceleration due to gravity = 9.81 m/s²
Therefore;
[tex]\displaystyle 6.0 \, s= \mathbf{\frac{u \cdot sin(45^{\circ})} {9.81}}[/tex]
Which gives;
[tex]\displaystyle The \ initial \ velocity \ of \ the \ cannonball, \ u = \frac{6.0 \, s \times 9.81 \, m/s^2}{sin(45^{\circ})} = \mathbf{58.86 \cdot \sqrt{2} \ m/s}[/tex]
According to law of conservation of linear momentum, we have;
M₁·v₁ₓ = m₂·v₂ₓ + m₃·v₃ₓ
v₁ₓ = The initial horizontal velocity = 58.86·√2 × cos(45°) m/s = 58.86 m/s
The magnitude of the smaller is the same as the initial velocity of the cannonball, given that the ball returns to point it was originally launched.
However, the direction of the smaller cannon ball is reversed.
Therefore, we have;
[tex]\displaystyle m \times 58.86 = \mathbf{\displaystyle \frac{2}{3} \times m\times v_{3x} - \displaystyle \frac{1}{3} \times m \times 58.86}[/tex]
Which gives;
[tex]\displaystyle 58.86 = \displaystyle \frac{2}{3} \times v_{3x} - \displaystyle \frac{1}{3} \times 58.86[/tex]
[tex]\displaystyle v_{3x} = \left( 58.86 + \displaystyle \frac{1}{3} \times 58.86\right) \times \frac{3}{2} = 117.72[/tex]
The horizontal velocity of the larger piece after the explosion, v₃ₓ = 117.72 m/s.
Given that the larger piece is also in free fall motion, it takes the same time, [tex]t_{max}[/tex] to land on the ground from the top of the trajectory (maximum height).
The distance from the launch site travelled by the larger piece, d, is given as follows;
[tex]d = \mathbf{v_{1x} \times t_{max} + v_{3x} \times t_{max}}[/tex]
Therefore;
d = 58.86 × 6 + 117.72 × 6 = 1059.48
The distance from the launch site the larger piece travels, d = 1,059.48 meters
Learn more about the conservation of linear momentum principle and projectiles here:
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