Question
on a air track, a 400g glider moving to the right at 2.00m/s collides elastically with a 500g glider moving in the opposite direction at 3.00 m/s.
Find the velocity of the first glider after the collision
Find the velocity of the second glider after the collision
Find the velocity of the first glider after the collision
Find the velocity of the second glider after the collision
Answers
The results will depend upon whether the collision is head-on or not. You have not provided that information.
In any case, total momentum and kinetic energy will conserved.
In any case, total momentum and kinetic energy will conserved.
Well, I do not have an answer, but lets say it IS a head on collision. How would this problem get solved?
M1 = 0.40kg, V1 = 2m/s.
M2 = 0.50kg, V2 = -3m/s.
Conservation of Momentum:
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.40*2 + 0.50*(-3) = 0.40*V3 + 0.50*V4,
Eq1: 0.4V3 + 0.5V4 = -0.70,
.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2(0.4-0.5) + 1.0*(-3))/(0.4+0.5)= -3.56 m/s. = Velocity of M1.
In Eq1, replace V3 with (-3.56) and solve for V4.
M2 = 0.50kg, V2 = -3m/s.
Conservation of Momentum:
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.40*2 + 0.50*(-3) = 0.40*V3 + 0.50*V4,
Eq1: 0.4V3 + 0.5V4 = -0.70,
.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2(0.4-0.5) + 1.0*(-3))/(0.4+0.5)= -3.56 m/s. = Velocity of M1.
In Eq1, replace V3 with (-3.56) and solve for V4.
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