Problem 545 - JEE Physics

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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

Step 5: 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.

Key Concepts

Impulse and Momentum

Impulse and momentum are closely related concepts in physics. Impulse refers to the change in momentum resulting from a force applied over a specific period of time. Think of it as a push that can change the motion of an object. For example, when the heavier block received an impulse, it gained velocity.
When discussing momentum, it is important to remember that it's a product of mass and velocity. Mathematically, momentum is expressed as \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

In the given exercise, the heavier block starts at rest, so its initial momentum is zero. After receiving an impulse that increases its velocity to \(14\,\mathrm{m/s}\), its momentum becomes \(140\,\mathrm{kg\cdot m/s}\) \( (10\,\mathrm{kg} \times 14\,\mathrm{m/s}).\) This change confirms impulse's role in altering momentum. Because the total initial momentum was zero, the impulse creates a total momentum of \(140\,\mathrm{kg\cdot m/s},\) applying the conservation of momentum in the system.

Center of Mass

The center of mass is an important concept in understanding the motion of composite systems, like two blocks connected by a spring. It represents the average position of all the mass in the system.

For the two blocks on a frictionless surface, the center of mass moves based on both blocks' masses and velocities. The formula to find the velocity of the center of mass \( (v_{cm}) \) is:\[v_{cm} =\frac{m_1v_1 + m_2v_2}{m_1 + m_2}\]Here, \( m_1 \) and \( m_2 \) are the masses of the two blocks, while \( v_1 \) and \( v_2 \) are their velocities. In the exercise, the heavier block has a mass of \(10\,\mathrm{kg}\) and moves at \(14\,\mathrm{m/s},\) while the lighter block has a mass of \(4\,\mathrm{kg}\) and remains at rest.

By plugging these values into the formula, the velocity of the center of mass is calculated to be \(10\,\mathrm{m/s}.\) This result illustrates that the combined motion of the system's center is influenced mainly by the moving heavier block, as the lighter block's inertia keeps its position unchanged.

Velocity Calculation

Velocity calculation is a fundamental aspect of understanding motion in physics, especially when different masses are in play. In the exercise problem, determining the velocity of the center of mass requires careful attention to both the individual velocities and the masses of the two blocks.

To calculate the velocity of the center of mass, we use the following formula:\[v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}\]Here, \(m_1 = 10\,\mathrm{kg}, m_2 = 4\,\mathrm{kg}, v_1 = 14\,\mathrm{m/s},\) and \(v_2 = 0\,\mathrm{m/s}.\) As you can see, the lighter block's velocity does not contribute due to it being zero.

By substituting these into our formula, the calculation simplifies:- First, multiply the mass and velocity for each block: \( (10\,\mathrm{kg} \times 14\,\mathrm{m/s} = 140\,\mathrm{kg\cdot m/s})\) and \( (4\,\mathrm{kg} \times 0\,\mathrm{m/s} = 0) \)- Divide the sum \( (140 + 0 = 140\,\mathrm{kg\cdot m/s})\) by the total mass \( (10 + 4 = 14\,\mathrm{kg}) \), yielding a result of \(10\,\mathrm{m/s}.\)

This value reflects the combined motion outcome of the two-block system, driven primarily by the moving heavy block. It shows how weighted averages are an effective way to determine the overall motion in physics.