Two blocks of masses \(10 \mathrm{~kg}\) an \(4 \mathrm{~kg}\) are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

Before giving the impulse to the heavier block, both blocks are at rest which means their velocities are 0. The total initial momentum of the system is 0.

After giving the impulse to the heavier block, it acquires a velocity of 14 m/s. Thus, its final momentum can be calculated as Final momentum of heavier block = mass (10 kg) × velocity (14 m/s) = 140 kg m/s.

Since the lighter block was not given any impulse, its momentum stays the same (initially 0). Therefore, the total final momentum of the system is the final momentum of the heavier block: Total final momentum = final momentum of heavier block = 140 kg m/s.

According to the conservation of momentum, the initial momentum of the system is equal to the final momentum of the system. Hence, we have: Initial momentum = Total final momentum 0 kg m/s = 140 kg m/s

The formula for the velocity of the center of mass is given by: \(v_{cm} =\frac{m_1v_1 + m_2v_2}{m_1 + m_2}\) Where \(m_1\) and \(m_2\) are the masses of the two blocks, and \(v_1\) and \(v_2\) are their respective velocities. In this case, \(m_1 = 10 \mathrm{~kg}\), \(m_2 = 4 \mathrm{~kg}\), \(v_1 = 14 \mathrm{~m/s}\), and \(v_2 = 0 \mathrm{~m/s}\). Plugging these values into the formula, we get: \(v_{cm} = \frac{(10 \mathrm{~kg})(14 \mathrm{~m/s}) +(4 \mathrm{~kg})(0 \mathrm{~m/s})}{(10 \mathrm{~kg})+(4 \mathrm{~kg})}\) \(v_{cm}= \frac{140 \mathrm{~kg} \cdot \mathrm{m/s}}{14 \mathrm{~kg}} = 10 \mathrm{~m/s}\) The velocity of the center of mass is \(10 \mathrm{~m/s}\). Therefore, the correct answer is option C.