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.
=
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.
5 answers
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.
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?
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.
c. 64.0
1 answer