Question

a block of mass m slides with velocity v on a frictionless horizontal surface and another block of mass 3m is at rest. the velocity of center of mass of system of two blocks is

a. v/5
b.v/4
c.5v/2
d.4v/5

Answer 1

Answer:

b. v/4

Explanation:

Mass of first block ($\rm m_1$) = m

Mass of second block ($\rm m_2$) = 3m

Velocity of first block ($\rm v_1$) = v

Velocity of second block ($\rm v_2$) = 0 (At Rest)

Velocity of center of mass of system of two blocks $= \rm \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \\\\ \rm = \dfrac{m \times v + 3m \times 0}{m + 3m} \\\\ \rm = \dfrac{\cancel{m}v}{4\cancel{m}} \\ \\ \rm = \dfrac{v}{4}$