Two blocks are sliding on a horizontal frictionless surface with velocities show

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 0.200 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?
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?
J

Explanation / Answer

a) It is an inelastic collision, so

Let the combined velocity be 'v'

v =m1v1+m2v2/(m1+m2)

v = (0.5*0.2 + 0.2*0.4)/ (0.5+0.2)

v = 0.257 m/s

b)

initial momentum = final momentum = p
p = 0.5kg * 0.2 m/s i + 0.2kg * 0.4m/s j = (0.1 i + 0.08 j) kg·m/s
The total mass is 0.7 kg, and
v = p / m = (0.142 i + 0.114 j) m/s

Direction is = arctan(0.114/0.142) = 38.82 degrees

c)

initial KE = ½ * 0.5kg * (0.2m/s)² + ½ * 0.2kg * (0.4m/s)² = 0.026 J
final KE = ½ * 0.7kg * (0.257m/s)² = 0.0231 J
Decrease is 2.88 *10^-3 J

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