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
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?
3 answers
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 = 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.