On a horizontal frictionless surface, a small block with mass 0.200 kg has a col

Question

On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediately after the collision, the 0.200 kg block is moving at 12.0 m/s in the direction 30° north of east and the 0.400 kg block is moving at 13.0 m/s in the direction 53.1° south of east. Use coordinates where the +x-axis is east and the +y-axis is north. (a) What is the total kinetic energy of the two blocks after the collision (in joules)? (b) What is the x-component of the total momentum of the two blocks after the collision? (Indicate the direction with the sign of your answer.) (c) What is the y-component of the total momentum of the two blocks after the collision? (Indicate the direction with the sign of your answer.)

Explanation / Answer

(a) total kinetic energy of the two blocks after the collision

KE = ½mv² = ½(0.2kg * (12m/s)² + 0.4kg * (13m/s)² = 48.2 J

b.)

What is the x-component of the total momentum of the two blocks after the collision?"
p_x = m*Vx = 0.2kg * 12m/s * cos30º + 0.4kg * 13m/s * cos-53.1º
p_x = 5.2 kg·m/s +x direction

c.) What is the y-component of the total momentum of the two blocks after the collision?"
p_y = m*Vy = 0.2kg * 12m/s * sin30º + 0.4kg * 13m/s * sin-53.1º
p_y = -2.958 kg·m/s in -y direction

(That is, "south.)

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