Answer: The velocity of the cart-2 is 2 m/s moving in direction opposite of the cart-1's direction.
Explanation:
Mass of cart-1 ,\(m_1=10.0 kg\)
Velocity of cart-1 before the release of the spring,\(u_1=0m/s\)
Mass of cart-2 ,\(m_1=15kg\)
Velocity of cart-2 before the release of the spring,\(u_2=0m/s\)
After releasing of the spring:
Velocity of cart-1 after the release of the spring,\(v_1=3 m/s\)
Velocity of cart-2 after the release of the spring,\(v_2=?\)
According to law of conservation of momentum:
\(m_1u_1+m_2u_2=m_2v_2+m_2v_2\)
\(10.0 kg\times 0m/s+15 kg\times 0 m/s=10.0 kg\times 3m/s+15v_2\)
\(0+0=30 kg m/s+15 mv_2\)
\(v_2=-2 m/s\)
Negative sign depicts that direction of cart-2 which is moving in the opposite direction of the cart-1's direction. So, the velocity of the cart 2 is 2 m/s.
