Question

Two gliders on an air track collide in a perfectly elastic collision. Glider A has mass 1.1 kg and is initially travelling at a velocity of 2.7 m/s [E]. It collides head-on with glider B with mass 2.4 kg, travelling at a velocity of 1.9 m/s [W]. Determine the final velocity of glider A using elastic collision formulas.

Answer 1

Conservation of Momentum:
M1*V1 + M2*V2 = M1*V3 + M2*V4.
1.1*2.7 + 2.4*(-1.9) = 1.1V3 + 2.4V4,
1.1V3 + 2.4V4 = -1.59.

Conservation of Kinetic Energy:
0.5M1*V1^2 + 0.5M2*V2^2 = 0.5M1*V3^2 + 0.5M2*V4^2.
Divide both sides by 0.5:
M1*V1^2 + M2*V2^2 = M1*V3^2 + M2*V4^2.
1.1*2.7^2 + 2.4*1.9^2 = 1.1V3^2 + 2.4V4^2,
1.1V3^2 + 2.4V4^2 = 16.7.

V3 = (V1(M1-M2)+2M2*V2)/(M1+M2).
V3 = (2.7(1.1-2.4)+4.8*(-1.9))/(1.1+2.4) = -3.61 m/s = 3.61 m/s, West = Final velocity of Glider A.























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Answer 2

Note: The kinetic energy Eq. derived was not required for this problem. It could have been used to calculate the total kinetic energy after the collision.

Answer 3

Two gliders collide on an air track. Glider 1 has a mass of 7.0 kg, and glider 2 has a mass of 4.0 kg. Before the collision, glider 1 had a velocity of 2.0 m/s, and glider 2 had a velocity of -5.0 m/s. If the collision is perfectly elastic, what is the total kinetic energy of both gliders after the collision?

Answer 4

consider the figure below. the two gliders are moving as shown in the figure and had a head on collision. if the velocity of A after impact is "-7.0" m/s. a) what is the velocity of B after collision? b) Determine the change in kinetic energy of the system.

Answer 5

c. 64.0