Question

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g glider at rest in the middle of the track. The end of the track over which the struck glider moves is not level, but slants upward at an angle of 0.70o with respect to the horizontal. Will the glider reach the end of the track? Neglect the length of the gliders.?

Answer 1

The glider will not reach the end of the track because the potential energy at the end of the track is less than the initial kinetic energy.

Step-by-step explanation:

Before we can determine whether the glider will reach the end of the track, we need to calculate the initial kinetic energy of the first glider and the potential energy at the end of the track. The initial kinetic energy is given by the formula KE = 1/2 * m * v^2, where m is the mass of the glider and v is its initial velocity. Plugging in the values, KE = 1/2 * 0.239 kg * (0.81 m/s)^2 = 0.0942 J.

Next, we need to calculate the potential energy at the end of the track. The potential energy is given by the formula PE = m * g * h, where m is the mass of the glider, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the track. Since the end of the track slants upward at an angle of 0.70 degrees, we can calculate the height using trigonometry. h = 2.4 m * sin(0.70 degrees) = 0.0297 m. Plugging in the values, PE = 0.239 kg * 9.8 m/s^2 * 0.0297 m = 0.0699 J.

Finally, we can compare the initial kinetic energy to the potential energy. Since the potential energy is less than the initial kinetic energy, the glider will not reach the end of the track.

Answer 2

Glider it stops just when it reaches the end of the runway

Step-by-step 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