A glider with a mass of 3 kg is moving rightward at a speed of 2 m/s on a frictionless horizontal air track. It collides with a stationary glider whose mass is 18 kg. If the collision is elastic, what is the speed of the gliders after the collision?
Use SI units in your answer.
velocity of the first glider =
Is the first glider moving left, right, or not moving?
velocity of the second glider =
Is the second glider moving left, right, or not moving?
momentum = 3 (2) + 18 (0) = 6 before and after
so
6 = 3 (u) + 18(v) or 2 = u + 6 v
so
u = (2 - 6 v) = 2 (1-3v)
Ke same before and after:
so
3(2^2) = 3u^2 + 18v^2
12 = 3(4)(1-3v)^2 + 18 v^2
2 = 2 (1 - 6v +9 v^2 ) + 3 v^2
0 = -12 v +18 v^2 + 3 v^2
21 v^2 = 12 v
21 v = 12
v = 12/21 = 4/7
check my arithmetic and go back and finish
Given:
M1 = 3kg, V1 = 2 m/s.
M2 = 18kg, V2 = 0.
M1*V1 + M2*V2 = M1*V4 + M2*V4.
3*2 + 18*0 = 3*V3 + 18*V4,
3*V3 + 18*V4 = 6,
Divide both sides by 3:
Eq1: V3 + 6*V4 = 2.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2(3-18) + 36*0)/(3+18) = -1.43 m/s = Velocity of M1 after collision.
In Eq1, replace V3 with -1.43 and solve for V4.
M1 is moving to the left(-) after collision.
To determine the speed of the gliders after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
The equation for conservation of momentum is:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Where:
m1 and m2 are the masses of the gliders,
v1_initial and v2_initial are the initial velocities of the gliders,
v1_final and v2_final are the final velocities of the gliders.
First, let's identify the initial velocities:
Given:
Mass of the first glider (m1) = 3 kg
Initial velocity of the first glider (v1_initial) = 2 m/s (moving rightward)
Mass of the second glider (m2) = 18 kg
Initial velocity of the second glider (v2_initial) = 0 m/s (stationary)
Now, substitute the known values into the equation:
(3 kg * 2 m/s) + (18 kg * 0 m/s) = (3 kg * v1_final) + (18 kg * v2_final)
Simplifying the equation:
6 kg * m/s = 3 kg * v1_final + 18 kg * v2_final
Since it is an elastic collision, both momentum and kinetic energy are conserved. Thus, we can also use the equation for conservation of kinetic energy:
(1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
Substituting the known values:
(1/2) * 3 kg * (2 m/s)^2 + (1/2) * 18 kg * (0 m/s)^2 = (1/2) * 3 kg * v1_final^2 + (1/2) * 18 kg * v2_final^2
Simplifying the equation:
6 J = (1/2) * 3 kg * v1_final^2 + 0 J
Since the velocity of the second glider (v2_initial) is zero, the velocity remains zero after the collision.
Now, we have two equations:
6 kg * m/s = 3 kg * v1_final + 18 kg * 0
6 J = (1/2) * 3 kg * v1_final^2 + 0 J
Solving the first equation for v1_final:
6 kg * m/s = 3 kg * v1_final
2 m/s = v1_final
So, the velocity of the first glider (v1_final) after the collision is 2 m/s (moving rightward).
As for the second glider, its velocity (v2_final) remains zero, as it was stationary initially.