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 0.350 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

given,

mass of A = 0.5 kg

mass of B = 0.35 kg

velocity of A = 0.2 m/s

velocity of B = 0.4 m/s

initial momentum = 0.5 * 0.2i + 0.35 * 0.4j

final momentum = (0.5 + 0.35) * v

by conservation of momentum

initial momentum = final momentum

0.5 * 0.2i + 0.35 * 0.4j = (0.5 + 0.35) * v

v = 0.1176i + 0.1647j

|v| = sqrt(0.1176^2 + 0.1647^2)

speed of the combined blocks = 0.202 m/s

angle = tan^-1(0.1647 / 0.1176)

angle does the velocity of the combined blocks make with the +x-axis = 54.47 degree

decrease in kinetic energy = initial kinetic energy - final kinetic energy

decrease in kinetic energy = 0.5 * 0.5 * 0.2^2 + 0.5 * 0.35 * 0.4^2 - 0.5 * (0.5 + 0.35) * 0.202^2

decrease in kinetic energy = 0.0206 J

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