Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi=0). (Figure 1)After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1>m2, so that after the collision, the two objects move off in the direction of the first object before the collision. What is the final speed vf of block 2?
Answers
The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]
the momentum is defined by the product of the mass and the velocity of the body.
p = mv
The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.
If the system is isolated, the forces during the collision are internal and the it is conserved. Let's find the momentum is two instants.
Initial instant. Before crash.
p₀ = m₁ u₀ + 0
Final moment. After crash.
[tex]p_f = m_1 u_f + m_2 v_f[/tex]
The momentum is preserved.
p₀ = [tex]p_f[/tex]
[tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]
Since the collision is elastic, the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex] + ½ m₂ [tex]v_f^2[/tex]
Let's write our system of equations.
[tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]
Let's solve
[tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]
[tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]
[tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
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Answer:
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What is the mechanical advantage?Mechanical advantage can be described as a measure of the ratio of output force to input force in a system. Mechanical advantage is used to analyze the forces in simple machines such as levers and pulleys.
Mechanical Advantage= output force (load) /input force(effort)
The mechanical advantage will increase as the slope of the inclined plane decreases but the load has to move with more distance. Under the mechanical advantage, the distance the load will move will be a fraction of the distance through applied force.
Thus, the mechanical advantage is decreased by some simple Machines decrease and are used to multiply distance to decrease speed.
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A 3.00 kg toy falls from a height of 10.0 m. Just before hitting the ground, what will its kinetic energy be? (Disregard air resistance; g = 9.81 m/s2.)
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The kinetic energy of the toy before hitting the ground is 294.3 J
From the question given above, the following data were obtained:
Mass (m) = 3 Kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.81 m/s²
Kinetic energy (KE) =?Before the toy will hit the ground, the potential and kinetic energy of the toy will be the same.
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Explanation:
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What is acceleration?Acceleration can be defined as the rate of change of velocity.
The acceleration can be calculated using the formula below.
Formula:
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Answer:
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Một oto đang chuyển động thẳng đều với vận tốc 36km/h thì hãm phanh sau 20s thì ô tô dừng lại quãng đường oto đi được là
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ASAP
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Given:
Q = - 622 kJ ( released of heat)W = 105 kJ.Substitute these values into equation 1
ΔU = -622-105ΔU = -727 kJHence, the change in internal energy of the system when it released a heat of 622 kJ and does a work of 105 kJ on the surroundings is -727 kJ.
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Explanation:
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