A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller diagram below.
1) Find the total energy for the roller coaster at the initial point.
2) Find the potential energy at point A using the PE formula.
3) Use the conservation of energy to find the kinetic energy (KE) at point B.
4) Find the potential energy at point C.
5) Use the conservation of energy to find the Kinetic Energy (KE) of the roller coaster at point C.
6) Use the Kinetic Energy from C, find velocity of the roller coaster at point C.
Answers
For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the initial point is 19820 J
2) The potential energy at point A is 19620 J
3) The kinetic energy at point B is 10010 J
4) The potential energy at point C is zero
5) The kinetic energy at point C is 19820 J
6) The velocity of the roller coaster at point C is 19.91 m/s
1) The total energy for the roller coaster at the initial point can be found as follows:
[tex] E_{t} = KE_{i} + PE_{i} [/tex]
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The total energy is:
[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]
Hence, the total energy for the roller coaster at the initial point is 19820 J.
2) The potential energy at point A is:
[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]
Then, the potential energy at point A is 19620 J.
3) The kinetic energy at point B is the following:
[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]
[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]
Since
[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]
we have:
[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]
Hence, the kinetic energy at point B is 10010 J.
4) The potential energy at point C is zero because h = 0 meters.
[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]
5) The kinetic energy of the roller coaster at point C is:
[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]
[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]
Therefore, the kinetic energy at point C is 19820 J.
6) The velocity of the roller coaster at point C is given by:
[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]
[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]
Hence, the velocity of the roller coaster at point C is 19.91 m/s.
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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?
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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]
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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.
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Final moment. After crash.
[tex]p_f = m_1 u_f + m_2 v_f[/tex]
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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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if you put 10 your right