Block 1, of mass m1 = 9.10 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mass m2 = 13.0 kg , which was initially at rest. The blocks stick together after the collision.
What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?
Express your answer numerically in joules.
Mass of block 1 = m1 = 9.1 kg
Mass of block 2 = m2 = 13 kg
Speed of block 1 before the collision = V1 = 27 m/s
Speed of block 2 before the collision = V2 = 0 m/s (At rest)
Speed of the blocks after the collision = V3
By conservation of linear momentum,
m1V1 + m2V2 = (m1 + m2)V3
(9.1)(27) + (13)(0) = (9.1 + 13)V3
V3 = 11.117 m/s
Initial kinetic energy of the system = KE1
KE1 = m1V12/2 + m2V22/2
KE1 = (9.1)(27)2/2 + (13)(0)2/2
KE1 = 3316.95 J
Final kinetic energy of the system = KE2
KE2 = (m1 + m2)V32/2
KE2 = (9.1 + 13)(11.117)2/2
KE2 = 1365.64 J
Change in kinetic energy of the system = KE
KE = KE2 - KE1
KE = 1365.64 - 3316.95
KE = -1951.31 J
Negative as energy is lost in the collision.
Change in the kinetic energy of the system due to the collision = -1951.31 J
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