Respuesta :
Answer:
Explanation:
In the elastic collision , law of conservation of momentum will be followed.
Initial total momentum of the system
= 123 u + 0 = 123u ( u is initial velocity of nate)
Final momentum of the system
= 899 x 1.61 + 123 v
= 1447.39 + 123 v
For conservation of momentum
123u = 1447.39 + 123 v
For perfectly elastic collision
relative velocity of approach = relative velocity of separation
u - 0 = 1.61 - v
Putting this value of u in the previous equation
123 ( 1.61 -v ) = 1447.39 +123v
198.03 - 123v = 1447.39 +123v
- 1249.36 = 246 v
v = - 5.078 m /s
u = 1.61 + 5.078
= 6.688 m /s
The momentum conservation allows to find the result for Nate's velocity in the elastic collision are:
- The initial velocity is: v₀ = 6.68 m / s
- The final velocity is: [tex]v_f[/tex] = - 5.08 m / s
Parameters given
- Nate's most with the skateboard m = 123 kg
- The mass of the statue M = 899 kg.
- The final velocity of the statue v =, 61 m / s
- Elastic collision
To find
- Skateboard speeds and Nate
The momentun is defined by the product of the mass and the velocity of a body, it is an isolated system this momentum is conserved.
p = m v
Where the bold letters indicate vectors, p is the moment, m the mass and v the velocity
If we define the system formed by nate and the skateboard and the statue, the system is isolating and the forces during the collision are internal, therefore the momentum is conserved.
Initial instant. Before the Shock.
p₀ = m v₀ + M 0
p₀ = mv₀
Final moment. After the crash.
[tex]p_f = m v_f + M v[/tex]
The momentum is preserved.
p₀ = [tex]p_f[/tex]
m v₀ = m [tex]v_f[/tex] + M v
They indicate that the collision is elastic, therefore the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m v₀² = ½ m [tex]v_f^2[/tex] + ½ M v²
m v₀² = m [tex]v_f^2[/tex] + M v²
Let's write our system of equations.
mv₀ = m [tex]v_f[/tex] + M v
m v₀² = m [tex]v_f^2[/tex] + M v²
Let's solve
m ([tex]v_f + \frac{M}{m}\ v[/tex] ) ² = m [tex]v_f^2[/tex] + M v²
[tex]v_f^2+ 2 \frac{M}{m} \ v \ v_f + (\frac{M}{m})^2 v^2 = v_f^2 + \frac{M}{m} \ v^2[/tex]
[tex]v_f ( 2 \frac{M}{m} \ v) = \frac{M}{m} v^2 ( 1 - \frac{M}{m}) \\v_f = \frac{v}{2} \ ( 1- \frac{M}{m} )[/tex]
Let's calculate
[tex]v_f = \frac{1.61}{2} \ (1- \frac{899}{123} )[/tex]
[tex]v_f[/tex] = -5.08 m / s
The negative sign indicates that the body goes in the opposite direction to the initial one.
Let's find the initial velocity.
v₀ = [tex]v_f + \frac{M}{m} \ v[/tex]
v₀ = [tex]-5.08 \ + \frac{899}{123} \ 1.61[/tex]
v₀ = 6.69 m / s
In conclusion using the conservation of momentum we can find the result for Nate's velocity in the elastic collision they are:
- The initial velocity is: v₀ = 6.68 m / s
- The final velocity is: [tex]v_f[/tex] = - 5.08 m / s
Learn more here: brainly.com/question/2141713
