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Answer :
So, the horizontal and vertical components of the momentum of the railcar and cannon just after the shell is fired are 9243.12 Kgm/s and 11011.25 Kgm/s, respectively.
Given in the question
Mass of the Shell= 125 Kg
Total mass of the railcar, cannon, and shell = 5000 Kg
Speed of projection = 115 m/s
Angle of Projection = 50.0 Degrees
Now the mass of railcar, and cannon = Total Mass - Mass of the shell
Now the mass of the railcar and cannon = 5000 - 125 Kg
Now the mass of the railcar, and cannon = is 4875 Kg
At the time of firing, according to the law of conservation of momentum,
Momentum of Railcar and cannon = Momentum of the Shell
Momentum of Railcar and cannon = Mass of shell × Speed of shell
Momentum of Railcar and cannon = 125 × 115 Kgm/s
Momentum of Railcar and cannon =14375 Kgm/s
So the momentum of the Railcar and the cannon is 14375 Kgm/s
To find the Components of momentum
Horizontal Component = Momentum × CosФ
Vertical Component = Momentum × SinФ
Ф given in the question is = 50.0 Degrees
Cos(50) = 0.643
Sin(50) = 0.766
Put in the value, we get
Horizontal Component = 14375 × 0.643 = 9243.12 Kgm/s
Vertical Component = 14375 × 0.766 = 11011.25 Kgm/s
So the horizontal and vertical components of the momentum of the railcar and cannon just after the shell is fired are 9243.12 Kgm/s and 11011.25 Kgm/s, respectively.
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The cannon is set so that it will project a 125 kg projectile at 115 m/s at an angle of 50.0o above the horizontal (and in the positive x direction). If the mass of the railcar, cannon, and shell prior to being fired is 5000 kg, what are the horizontal and vertical components of the momentum of the railcar and cannon just after the shell is fired?
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Rewritten by : Barada
The horizontal component of the railcar and cannon's velocity just after the shell is fired is approximately [tex]\( 1.848 \, \text{m/s} \)[/tex], and the vertical component is approximately [tex]\( 2.2022 \, \text{m/s} \).[/tex]
To solve this problem, we will use the principle of conservation of momentum. Before the cannon is fired, the total momentum of the system (railcar, cannon, and projectile) is zero because everything is at rest. When the cannon is fired, the momentum of the projectile in both the horizontal and vertical directions must be equal in magnitude and opposite in direction to the momentum of the railcar and cannon combined, due to the conservation of momentum.
Let's denote the mass of the railcar and cannon as [tex]\( m_{rc} = 5000 \, \text{kg} \)[/tex] and the mass of the projectile as [tex]\( m_{p} = 125 \, \text{kg} \)[/tex]. The initial velocity of the projectile is [tex]\( v_{p} = 115 \, \text{m/s} \)[/tex] at an angle[tex]\( \theta = 50.0^{\circ} \)[/tex] above the horizontal.
First, we need to find the horizontal and vertical components of the projectile's initial velocity. The horizontal component \( v_{px} \) can be found using the cosine of the angle:
[tex]\[ v_{px} = v_{p} \cos(\theta) \][/tex]
[tex]\[ v_{px} = 115 \, \text{m/s} \times \cos(50.0^{\circ}) \][/tex]
[tex]\[ v_{px} \approx 115 \, \text{m/s} \times 0.6428 \][/tex]
[tex]\[ v_{px} \approx 73.92 \, \text{m/s} \][/tex]
The vertical component [tex]\( v_{py} \)[/tex] can be found using the sine of the angle:
[tex]\[ v_{py} = v_{p} \sin(\theta) \][/tex]
[tex]\[ v_{py} = 115 \, \text{m/s} \times \sin(50.0^{\circ}) \][/tex]
[tex]\[ v_{py} \approx 115 \, \text{m/s} \times 0.7660 \][/tex]
[tex]\[ v_{py} \approx 88.09 \, \text{m/s} \][/tex]
Now, let's denote the horizontal and vertical components of the railcar and cannon's velocity after the shell is fired as [tex]\( v_{rcx} \)[/tex] and [tex]\( v_{rcy} \),[/tex]respectively. According to the conservation of momentum:
For the horizontal direction:
[tex]\[ m_{rc} \cdot v_{rcx} = m_{p} \cdot v_{px} \][/tex]
[tex]\[ v_{rcx} = \frac{m_{p} \cdot v_{px}}{m_{rc}} \][/tex]
[tex]\[ v_{rcx} = \frac{125 \, \text{kg} \times 73.92 \, \text{m/s}}{5000 \, \text{kg}} \][/tex]
[tex]\[ v_{rcx} = \frac{9240 \, \text{kg} \cdot \text{m/s}}{5000 \, \text{kg}} \][/tex]
[tex]\[ v_{rcx} = 1.848 \, \text{m/s} \][/tex]
For the vertical direction:
[tex]\[ m_{rc} \cdot v_{rcy} = m_{p} \cdot v_{py} \][/tex]
[tex]\[ v_{rcy} = \frac{m_{p} \cdot v_{py}}{m_{rc}} \][/tex]
[tex]\[ v_{rcy} = \frac{125 \, \text{kg} \times 88.09 \, \text{m/s}}{5000 \, \text{kg}} \][/tex]
[tex]\[ v_{rcy} = \frac{11011.25 \, \text{kg} \cdot \text{m/s}}{5000 \, \text{kg}} \][/tex]
[tex]\[ v_{rcy} = 2.2022 \, \text{m/s} \][/tex]
The final answer is:
[tex]\[ \boxed{v_{rcx} = 1.848 \, \text{m/s}} \][/tex]
[tex]\[ \boxed{v_{rcy} = 2.2022 \, \text{m/s}} \][/tex]
