Complete question:
Block A on the left has a mass of 1.0 kg, Block B on the right has a mass of 3.0 kg.
Block A is initially moving to the right at 6.00 m/s, while block B is initially at rest. The surface they move on is level and frictionless. What is the velocity of the center of mass of the two blocks after the blocks collide
Answer:
The velocity of center mass of the two blocks after collision is 1.5 m/s to the right.
Explanation:
Let the mass of block A = m₁ = 1 kg
Let the mass of block B = m₂ = 3kg
initial velocity of block A, u₁ = 6 m/s
initial velocity of block B, u₂ = 0
let the velocity of center mass of the blocks after collision = v
Apply the principle of conservation of linear momentum for inelastic collision (the two blocks sticked together after collision);
[tex]m_1u_1 + m_2u_2 = v(m_1+m_2)\\\\m_1u_1 + m_2(0) = v(m_1+m_2)\\\\m_1u_1 = v(m_1+m_2)\\\\v = \frac{m_1u_1}{m_1+m_2}\\\\v= \frac{1*6}{1+3}\\\\v= \frac{6}{4}\\\\v= 1.5 \ m/s \ \ \ to \ the \ right[/tex]
v = 1.5 m/s to the right
Therefore, the velocity of center mass of the two blocks after collision is 1.5 m/s to the right.
