Answer:
x = 0.0756 m
Fred moves in the direction where Brutus moves
Explanation:
This exercise is for the moment, we define the system as formed by the two players, for this system the forces in the clash are internal, so the moment is preserved
Initial. Before the crash
p₀ = m v₀₁ - M v₀₂
Final. After the crash
= (m + M) v
p₀ = ![p_{f}](https://tex.z-dn.net/?f=p_%7Bf%7D)
m v₀₁ –M v₀₂ = (m + M) v
v = (m v₀₁ - M v₀₂) / (m + M)
Let's calculate
v = (60 6 - 120 4) / (60 +120)
v = - 120/180
v = - 0.667 m / s
The negative sign indicates that the final speed is the direction where Brutus runs
Let's use Newton's second law to find the acceleration of the two players
fr = (m + M) a
fr = μ N
N- W = 0
N = (m + M) g
μ (m + M) g = (m + M) a
a = μ g
a = 0.30 9.8
a = 2.94 m / s²
We use kinematics to find the distance traveled, the final speed is zero
v² = v₀² - 2 a x
x = v₀² / 2 a
x = 0.667² / (2 2.94)
x = 0.0756 m
PEG = 0.8*KE.
M*g*h = 0.8 *(0.5*M*v^2) M, h, and v are the vaulter's mass, height, and speed, respectively.
g*h = 0.8*0.5*v^2 Divide both sides by M (the mass).
g*h = 0.4*v^2
(9.81 m/s^2)*h = 0.4*(7 m/s)^2. Note that m and s denote meter and second, respectively.
h = 0.4*(7 m/s)^2 /(9.81 m/s^2) Divide both sides by 9.81 m/s^2. Also, note that m is NOT mass here.
= 0.4*(49 m^2/s^2)/(9.81 m/s^2)
= 0.4*49 m/9.81
Answer = 1.99796 meters
The horizontal distance traveled below the table's edge is 29.4 cm.
The given parameters;
- <em>mass of the bullet, = 15 g = 0.015 kg</em>
- <em>speed of the bullet, = 630 m/s</em>
- <em>mass of the block, = 4.9 kg</em>
The final velocity of the bullet-block system is calculated as follows;
![m_1 u_1 + m_2u_1 = v(m_1 + m_2)\\\\0.015(630) + 4.9(0) = v(0.015 + 4.9)\\\\9.45 = 4.915 v\\\\v = \frac{9.45}{4.915} \\\\v = 1.923 \ m/s](https://tex.z-dn.net/?f=m_1%20u_1%20%2B%20m_2u_1%20%3D%20v%28m_1%20%2B%20m_2%29%5C%5C%5C%5C0.015%28630%29%20%2B%204.9%280%29%20%3D%20v%280.015%20%2B%204.9%29%5C%5C%5C%5C9.45%20%3D%204.915%20v%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7B9.45%7D%7B4.915%7D%20%5C%5C%5C%5Cv%20%3D%201.923%20%5C%20m%2Fs)
The time for the bullet-block system to reach the ground from the table is calculated as follows;
![h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.75}{9.8} }\\\\t = 0.153 \ s](https://tex.z-dn.net/?f=h%20%3D%20v_0t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5C2h%20%3D%20gt%5E2%5C%5C%5C%5Ct%5E2%20%3D%20%5Cfrac%7B2h%7D%7Bg%7D%20%5C%5C%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D%20%5C%5C%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ctimes%200.75%7D%7B9.8%7D%20%7D%5C%5C%5C%5Ct%20%3D%200.153%20%5C%20s)
The horizontal distance traveled below the table's edge is calculated as follows;
![X = vt\\\\X = 1.923 \times 0.153\\\\X = 0.294 \ m\\\\X = 29.4 \ cm](https://tex.z-dn.net/?f=X%20%3D%20vt%5C%5C%5C%5CX%20%3D%201.923%20%5Ctimes%200.153%5C%5C%5C%5CX%20%3D%200.294%20%5C%20m%5C%5C%5C%5CX%20%3D%2029.4%20%5C%20cm)
Thus, the horizontal distance traveled below the table's edge is 29.4 cm.
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