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 0.450 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.

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 0.450 kg. The two blocks have a perfectly inelastic collision and stick together after the collision. What is the speed of the combined blocks after the collision? What angle does the velocity of the combined blocks make with the +x-axis after the collision? A A counterclockwise from the +x-axis What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Explanation / Answer

m1 v1 = (m1 + m2) Vx = M Vx

m2 v2 = (m1 + m2) Vy = M Vy

Vx = .5 * .2 / .95 = .105 m/s

Vy = .45 * .4 / .95 = .189 m/s

V = (.105^2 + .189^2)^1/2 = .217 m/s

tan theta = Vy / Vx = .189 / .105

theta = 60.9 deg above x-axis

KE (lost) = 1/2 m1 v1^2 + 1/2 m2 v2^2 - 1/2 M V^2

KE (lost) = 1/2 (.5 * .04 + .45 * .16 - .95 * .217^2) = .0236 J

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