Answer:
Glider it stops just when it reaches the end of the runway
Explanation:
This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy
Po = pf
Ko = Kf
Before crash
Po = m1 Vo1 + 0
Ko = ½ m1 Vo1²
After the crash
Pf = m1 Vif + Vvf
Kf = ½ m1 V1f² + ½ m2 V2f²
m1 V1o = m1 V1f + m2 V2f (1)
m1 V1o² = m1 V1f² + m2 V2f² (2)
We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2
m1 (V1o -V1f) = m2 V2f (3)
m1 (V1o² - V1f²) = m2 V2f²
Let's use the relationship (a + b) (a-b) = a² -b²
m1 (V1o + V1f) (V1o -V1f) = m2 V2f²
We divide with 3 and simplify
(V1o + V1f) = V2f (4)
Substitute in 3, group and clear
m1 (V1o - V1f) = m2 (V1o + V1f)
m1 V1o - m2 V1o = m2 V1f + m1 V1f
V1f (m1 -m2) = V1o (m1 + m2)
V1f = V1o (m1-m2 / m1+m2)
We substitute in (4) and group
V2f = V1o + (m1-m2 / m1 + m2) V1o
V2f = V1o [1+ + (m1-m2 / m1 + m2)]
V2f = V1o (2m1 / (m1+m2)
We calculate with the given values
V1f = 0.81 (239-513 / 239 + 513)
V1f = 0.81 (-274/752)
V1f = - 0.295 m/s
The negative sign indicates that the planned one moves in the opposite direction to the initial one
V2f = 0.81 [2 239 / (239 + 513)]
V2f = 0.81 [0.636]
V2f = 0.515 m / s
Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,
Eo = Ef
Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point
Eo = K = ½ m2 vf2²
Ef = U = m2 g Y
½ m2 v2f² = m2 g Y
Y = V2f² / 2g
Y = 0.515²/2 9.8
Y = 0.0147 m
At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track
tan θ = Y / x
Y = x tan θ
The crash occurs in the middle of the track whereby x = 1.2 m
Y = 1.2 tan 0.7
Y = 0.147 m
As the two quantities are equal in glider it stops just when it reaches the end of the runway
