Question

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 : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

Short answer

The velocity of the center of mass is \(10 \mathrm{~m/s}\). Therefore, the correct answer is option C.

Step-by-step solution

Calculate the initial momentum

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.

Calculate the final momentum of the heavier block

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.

Calculate the total final momentum of the system

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.

Apply conservation of momentum

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

Find the velocity of the center of mass

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.