Answer to Question #189363 in Mechanics | Relativity for Ayesha

Question

Two blocks of mass 0.1 kg and 0.2 kg approach each other on a horizontal plane at velocities of 0.4 and 1 m/s respectively. If the blocks collide and remain together, calculate the joint velocity after collision.

Answer 1

The total momentum before the collision and after the collision is conserved

$p_1+p_2 = p_3$

$m_1v_1+m_2v_2=m_3v_3$

$\displaystyle v_3 = \frac{m_1v_1+m_2v_2}{m_3} = \frac{0.1 \cdot 0.4 - 0.2 \cdot 1}{0.1+0.2} = -0.533$ m/s.

The minus sign in the formula above appears because the velocity of the blocks had different directions. The minus sign of $v_3$ just means that the joint block will move in the direction of the second block.

Answer: 0.533 m/s.