Question

Illustration - 14 Two 100 gm blocks hang at the ends of a light flexible cord passing over a small frictionless pulley. A 40 gm block rests on the block on right and removed after 2 seconds. (a) How far will each block move in the first second after 40 gm block is removed? (b) What was the tension in the cord before the 40 gm block was removed? (c) What was the tension in the cord supporting the pulley after 40 gm block was removed?

• A. (a) Each block will move $\frac{9.8}{3}$ meters. The block on the left moves upwards, and the block on the right moves downwards. (b) The tension was $\frac{3.43}{3}$ N. (c) The tension supporting the pulley was $1.96$ N.
• B. (a) Each block will move $1.63$ meters. (b) The tension was $1.14$ N. (c) The tension supporting the pulley was $0.98$ N.
• C. (a) Each block will move $3.27$ meters. (b) The tension was $1.14$ N. (c) The tension supporting the pulley was $1.96$ N.
• D. (a) Each block will move $\frac{9.8}{3}$ meters. (b) The tension was $1.14$ N. (c) The tension supporting the pulley was $0.98$ N.

Answer 1

Answer: (A)

(a) Each block will move $\frac{9.8}{3}$ meters. The block on the left moves upwards, and the block on the right moves downwards. (b) The tension was $\frac{3.43}{3}$ N. (c) The tension supporting the pulley was $1.96$ N.

Answer 2

The problem involves two phases of motion for a system of blocks connected by a cord over a pulley. We assume $g = 9.8 \text{ m/s}^2$.

Phase 1: First 2 seconds The masses involved are $m_1 = 100$ gm $= 0.1$ kg (left block) and $m_R = 100 + 40 = 140$ gm $= 0.14$ kg (right block with the 40 gm block on top). The acceleration of the system ($a_1$) is calculated using Newton's second law for the combined system: $a_1 = \frac{m_R - m_1}{m_R + m_1} g = \frac{0.14 \text{ kg} - 0.1 \text{ kg}}{0.14 \text{ kg} + 0.1 \text{ kg}} g = \frac{0.04}{0.24} g = \frac{1}{6} g$. Substituting $g = 9.8 \text{ m/s}^2$, we get $a_1 = \frac{9.8}{6} = \frac{4.9}{3} \text{ m/s}^2$. The velocity at the end of 2 seconds ($v(2)$), assuming the system starts from rest, is $v(2) = a_1 \times 2 = \frac{4.9}{3} \times 2 = \frac{9.8}{3} \text{ m/s}$.

Phase 2: After 2 seconds The 40 gm block is removed, so the masses on each side are now equal: $m_1 = 0.1$ kg and $m_2 = 0.1$ kg. Since the masses are equal, the net force on the system is zero, and thus the acceleration $a_2 = 0$. The velocity remains constant at $v(t) = v(2) = \frac{9.8}{3} \text{ m/s}$ for $t > 2$ s.

(a) Displacement in the first second after removal: This refers to the displacement between $t=2$s and $t=3$s, so $\Delta t = 1$ s. The initial velocity for this interval is $v(2) = \frac{9.8}{3} \text{ m/s}$, and the acceleration is $a_2 = 0$. The displacement ($\Delta y$) is given by $\Delta y = v(2) \Delta t + \frac{1}{2} a_2 (\Delta t)^2$. $\Delta y = \left(\frac{9.8}{3} \text{ m/s}\right) \times (1 \text{ s}) + \frac{1}{2} \times (0 \text{ m/s}^2) \times (1 \text{ s})^2 = \frac{9.8}{3}$ meters. Each block will move $\frac{9.8}{3}$ meters. The left block moves upwards, and the right block moves downwards.

(b) Tension before removal: This is the tension ($T_1$) during Phase 1. Considering the left block ($m_1$): $T_1 - m_1 g = m_1 a_1$ $T_1 = m_1 (g + a_1) = 0.1 \text{ kg} \left(9.8 \text{ m/s}^2 + \frac{9.8}{6} \text{ m/s}^2\right) = 0.1 \times 9.8 \times \left(1 + \frac{1}{6}\right) = 0.98 \times \frac{7}{6} = \frac{6.86}{6} = \frac{3.43}{3}$ N.

(c) Tension supporting the pulley after removal: This is the tension in the cord holding the pulley during Phase 2. Since $a_2 = 0$, the tension on each side of the pulley is equal to the weight of the block on that side: $T_2 = m_1 g = m_2 g$. $T_2 = 0.1 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.98$ N. The cord supporting the pulley bears the tension from both sides: $T_{pulley} = T_2 + T_2 = 0.98 \text{ N} + 0.98 \text{ N} = 1.96$ N.