Two particles of mass m_1 = 0.260 kg and m_2 = 0.150 kg collide on a smooth hori

Question

Two particles of mass m_1 = 0.260 kg and m_2 = 0.150 kg collide on a smooth horizontal surface as shown in the diagram. a) Determine v_2f (the final velocity of m_2) if v_1, = 0.540 m/s, v_2 = 0.900 m/s, and v_1, = 0.700 m/s. b) Is the collision elastic? If not, calculate the fractional change in the kinetic energy (Delta K/K_i = k_f - k_i/k_1). Numerical values are required A machinist turns on the power to a grinding wheel at time t = 0 s. The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 38 rad/s. The wheel is run at that angular speed for 30 s and then power is shut off. The wheal

Explanation / Answer

v1i = 0.54 i

v2i = 0.9*sin50i - 0.9*cos50 j = 0.689i - 0.578 j

v1f = 0.7*cos35i - 0.7*sin35j = 0.573i - 0.401 j

v2f = ?

from conservation of momentum

momentum before collision = momentum aftercollision

m1*v1i + m2*v2i = m1*v1f + m2*v2f

(0.26*0.54i) + 0.15*( 0.689i - 0.578 j) = (0.26*(0.573i - 0.401 j)) + (0.15*v2f)

v2 = 0.632i + 0.117 j

magnitude = sqrt(0.632^2+0.117^2) = 0.642 m/s

direction = tan^-1(0.117/0.632) = 10.5 degrees

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initial kinetic energy Ki = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2

Ki = (1/2)*0.26*0.54^2 + (1/2)*0.15*0.9^2 = 0.0987 J

final kinetic energy Kf = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2

Kf = (1/2)*0.26*0.7^2 + (1/2)*0.15*0.632^2 = 0.0936 J

Kf < Ki

collision is not elastic

dK/K = (0.0936-0.0987)/0.0987 = -0.516

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