On an air track, a 403.0 glider moving to the right at 2.30 collides elastically with a 501.0 glider moving in the opposite direction at 3.00 .Find the velocity of first glider after the collision
Given:
M1 = 0.403kg, V1 = 2.30 m/s.
M2 = 0.501kg, V2 = -3.00 m/s.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2.3(0.403-0.501)+1.002*(-3.0))/0.904 = -3.57 m/s = Velocity of M1 after collision.
V3 is moving to the left.
To find the velocity of the first glider after the collision, we can use the principle of conservation of momentum.
The conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. In this case, the system consists of the two gliders.
The momentum of an object is defined as the product of its mass and velocity. Mathematically, momentum (p) can be calculated as p = m * v, where p is the momentum, m is the mass, and v is the velocity.
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it can be written as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where m1 and m2 are the masses of the gliders, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.
In this case, let's assume the first glider (m1 = 403.0 g) initially moves to the right at a velocity of v1 = 2.30 m/s, while the second glider (m2 = 501.0 g) moves to the left at v2 = -3.00 m/s.
Using the conservation of momentum equation, we can substitute these values and solve for v1':
(403.0 g * 2.30 m/s) + (501.0 g * -3.00 m/s) = (403.0 g * v1') + (501.0 g * v2')
Simplifying the equation, we get:
(926.9 g*m/s) - (1503.0 g*m/s) = (403.0 g * v1') - (501.0 g * 3.00 m/s)
-576.1 g*m/s = (403.0 g * v1') - 1503.0 g*m/s
Rearranging and solving for v1', we have:
403.0 g * v1' = -576.1 g*m/s + 1503.0 g*m/s
403.0 g * v1' = 926.9 g*m/s
v1' = 926.9 g*m/s / 403.0 g
v1' ≈ 2.299 m/s
Therefore, the velocity of the first glider after the collision is approximately 2.299 m/s to the right.