Block 1 (mass = m_1 = M) slides at a speed of v_i = v in the +x direction toward

Question

Block 1 (mass = m_1 = M) slides at a speed of v_i = v in the +x direction towards block 2 (mass = m_2 = 2m) which is at rest. The surface is frictionless and horizontal. Assuming that the blocks collide and stick together, what percent of the initial system kinetic energy is lost? a. 75% b. 67% c. 25% d. 33% e. 100% At left, three particles (1,2,3) having masses: M_1 = 20 kg, M_2 = 10 kg, and M_3 = 30 kg are fixed on the xy plane. Four points (a right arrow D) are indicated on the figure. Which one of the four points comes closest to the position of the center of mass of the system? a. A b. B c. C d. D e. not enough information provided to answer the question!!!!!

Explanation / Answer

6)

we use conservation of momentum

M*v + 2M*0 = (M + 2M)*vf

vf = v/3

kinetic energy before the collision = (1/2)Mv^2

kinetic energy after the collision = (1/2)*3M*(v/3)^2 = (1/3)*[(1/2)Mv^2]

lost kinetic energy = (1/2)Mv^2 - (1/3)*[(1/2)Mv^2] = (2/3)*[(1/2)Mv^2]

percentage of lost kinetic energy = [ (2/3)*[(1/2)Mv^2] / (1/2)Mv^2 ]*100 = 67 %

7)

let L = 1 m

x component of CoM = (20*0 + 10*0 + 30*1)/(20+10+30) = (1/2)m

y component of CoM = (20*1 + 10*0 + 30*1/2)/(20+10+30) = (7/12) m

CoM = (0.5L, 0.58L)

now we can see

D is closest.

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