Asked by lorena
A 0.400 glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800 when it makes a head-on collision with a stationary 0.150 glider.Find the magnitude of the final velocity of first glider if the collision is elastic. Find the direction of the final velocity of first glider if the collision is elastic.
Find the magnitude of the final velocity of second glider if the collision is elastic. Find the direction of the final velocity of second glider if the collision is elastic. Find the final kinetic energy of first glider. Find the final kinetic energy of second glider.
Find the magnitude of the final velocity of second glider if the collision is elastic. Find the direction of the final velocity of second glider if the collision is elastic. Find the final kinetic energy of first glider. Find the final kinetic energy of second glider.
Answers
Answered by
drwls
No dimension -- no help
Answered by
Henry
M1 = 0.400kg, V1 = 0.800 m/s.
M2 = 0.150kg, V2 = 0.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.4*0.8 + 0.15*0 = 0.4*V3 * 0.15*V4,
Eq1: 0.4*V3 + 0.15*V4 = 0.32.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0.8(0.4-0.15) + 0.30*0)/(0.4+0.15) = 0.20 /0.55 = 0.364 m/s,rt. = Velocity of M1 after the collision.
In Eq1, replace V3 with 0.364 and solve for V4:
0.4*0.364 + 0.15*V4 = 0.32.
V4 = 1.16 m/s,rt. = Velocity of M2 after the collision.
KE1 = 0.5*M1*V3^2.
KE2 = 0.5*M2*V4^2.
M2 = 0.150kg, V2 = 0.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.4*0.8 + 0.15*0 = 0.4*V3 * 0.15*V4,
Eq1: 0.4*V3 + 0.15*V4 = 0.32.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0.8(0.4-0.15) + 0.30*0)/(0.4+0.15) = 0.20 /0.55 = 0.364 m/s,rt. = Velocity of M1 after the collision.
In Eq1, replace V3 with 0.364 and solve for V4:
0.4*0.364 + 0.15*V4 = 0.32.
V4 = 1.16 m/s,rt. = Velocity of M2 after the collision.
KE1 = 0.5*M1*V3^2.
KE2 = 0.5*M2*V4^2.
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