Two blocks are sliding on a horizontal frictionless surface with velocities show
ID: 1425532 • Letter: T
Question
Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.500 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.
a) What is the speed of the combined blocks after the collision?
in m/s
(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?
° counterclockwise from the +x-axis
(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?
in J
Explanation / Answer
initial momentum = final momentum = p
p = 0.5kg * 0.2 m/s i + 0.5kg * 0.4m/s j = (0.1 i + 0.1 j) kg·m/s
The total mass is 0.75 kg, and
v = p / m = (0.13 i + 0.13 j) m/s
(a) Then speed = (0.13² + 0.13²) m/s = 0.189 m/s
(b) = 45º (by observation, or by = arctan(0.13/0.13)
(c) initial KE = ½ * 0.5kg * (0.2m/s)² + ½ * 0.5kg * (0.4m/s)² = 0.03 J
final KE = ½ * 0.75kg * (0.189m/s)² = 0.0133 J
Decrease is 16.7 mJ
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