Question

Two masses A & B each of 5 kg are suspended by a light inextensible string passing over a smooth massless pulley such that mass A rest on smooth table & B is held at the position shown. Mass B is now gently lifted up to the pulley and allowed to fall from rest. Determine up to what height will A rise for the ensuing motion.

Answer 1

0.00

Answer 2

The problem describes a system with two masses, A and B, each of 5 kg, connected by a light inextensible string passing over a smooth massless pulley. Mass A rests on a smooth horizontal table, and mass B hangs vertically.

The problem states that "Mass B is now gently lifted up to the pulley and allowed to fall from rest."

Stage 1: Motion of A and B until B hits the ground.

As mass B falls, mass A moves horizontally on the smooth table. Since the string is inextensible and the pulley is massless and smooth, the speed of A will be equal to the speed of B, and the distance A moves horizontally will be equal to the distance B falls.

We can use the principle of conservation of mechanical energy for this stage, as the table is smooth (no friction) and the pulley is massless and smooth (no energy loss).

Let $m_A = m_B = m = 5 \text{ kg}$. Let $g = 9.8 \text{ m/s}^2$.

Initial state (B at pulley, A at rest):

• Potential Energy ($PE_i$): We can set the ground level as the reference for potential energy. So, $PE_i = m_B g h_1 = m g h_1$.
• Kinetic Energy ($KE_i$): Both masses are at rest, so $KE_i = 0$.

Final state (B at ground, A and B moving with velocity $v$):

• Potential Energy ($PE_f$): Mass B is at the ground level, so its potential energy is 0. Mass A remains on the table, so its potential energy relative to the table does not change (it's always 0 if the table is the reference). So, $PE_f = 0$.
• Kinetic Energy ($KE_f$): Both masses are moving with velocity $v$. So, $KE_f = \frac{1}{2} m_A v^2 + \frac{1}{2} m_B v^2 = \frac{1}{2} (m+m) v^2 = \frac{1}{2} (2m) v^2 = m v^2$.

By conservation of mechanical energy: $PE_i + KE_i = PE_f + KE_f$ $m g h_1 + 0 = 0 + m v^2$ $m g h_1 = m v^2$ $v^2 = g h_1$

Stage 2: Motion of A after B hits the ground.

When mass B hits the ground, it comes to rest (assuming it does not bounce). At this point, the string connecting A and B becomes slack.

Mass A is on a smooth horizontal table and has acquired a velocity $v = \sqrt{19.6} \text{ m/s}$.

Since mass A is on a smooth horizontal table and the string is now slack (because B has stopped), there are no horizontal forces acting on A (it's smooth, so no friction). There are also no vertical forces that would cause it to rise. The normal force from the table balances its weight. Therefore, mass A will continue to move horizontally with the constant velocity $v$ it acquired. It will not rise vertically.

Thus, the height A will rise is 0 m.