Question

Two particles A and B of mass 2m and m respectively are attached to the ends of a light inextensible string of length 4a which passes over a small smooth peg at a height 3a from an inelastic table. The system is released from rest with each particle at a height a from the table. Find

(i) The speed of B when A strikes the table. (ii) The time that elapses before A first hits the table. (iii) The time for which A is resting on the table after the first collision & before it is first jerked off.

Answer 1

i) $\sqrt{\frac{2ga}{3}}$ ii) $\sqrt{\frac{6a}{g}}$ iii) $\sqrt{\frac{6a}{g}}$

Answer 2

The problem describes a system of two masses connected by a string over a pulley. We need to analyze its motion in three parts.

System Setup:

• Mass of particle A ($m_A$) = 2m
• Mass of particle B ($m_B$) = m
• Length of string = 4a
• Height of the peg from the table = 3a
• Initial height of each particle from the table = a
• The system is released from rest.

Analysis of Initial State and Motion: Since both particles are initially at height 'a' and the peg is at '3a', the length of the string on A's side is (3a - a) = 2a, and on B's side is (3a - a) = 2a. The total string length is 2a + 2a = 4a, which is consistent with the given information. When particle A strikes the table, it falls a distance of 'a'. Consequently, particle B rises a distance of 'a'.

1. Calculate the acceleration of the system: Let T be the tension in the string and $a'$ be the acceleration of the system. For particle A (moving downwards): $m_A g - T = m_A a'$ $2mg - T = 2m a' \quad (1)$

For particle B (moving upwards): $T - m_B g = m_B a'$ $T - mg = m a' \quad (2)$

Adding (1) and (2): $(2mg - T) + (T - mg) = 2m a' + m a'$ $mg = 3m a'$ $a' = \frac{g}{3}$

Part (i): The speed of B when A strikes the table. Particle A falls a distance $s = a$. The initial velocity $u = 0$. Using the kinematic equation $v^2 = u^2 + 2a's$: $v^2 = 0^2 + 2 \left(\frac{g}{3}\right) (a)$ $v^2 = \frac{2ga}{3}$ $v = \sqrt{\frac{2ga}{3}}$

Part (ii): The time that elapses before A first hits the table. Particle A falls a distance $s = a$. Initial velocity $u = 0$. Using the kinematic equation $s = ut + \frac{1}{2}a't^2$: $a = 0 \cdot t + \frac{1}{2} \left(\frac{g}{3}\right) t^2$ $a = \frac{g}{6} t^2$ $t^2 = \frac{6a}{g}$ $t = \sqrt{\frac{6a}{g}}$

Part (iii): The time for which A is resting on the table after the first collision & before it is first jerked off.

• Moment A strikes the table:

  • A comes to rest (velocity = 0).
  • B is moving upwards with velocity $v = \sqrt{\frac{2ga}{3}}$.
  • B's height from the table is now $a + a = 2a$.
  • At this instant, the string becomes slack as A is no longer pulling it.

• Motion of B after A hits the table:

  • B continues to move upwards under gravity with initial velocity $v = \sqrt{\frac{2ga}{3}}$ and acceleration $a_B = -g$.
  • Time for B to reach its highest point ($t_1$): Using $v_f = v_i + a_B t_1$: $0 = \sqrt{\frac{2ga}{3}} - g t_1$ $t_1 = \frac{1}{g} \sqrt{\frac{2ga}{3}} = \sqrt{\frac{2a}{3g}}$
  • Height B rises further ($s_1$): Using $v_f^2 = v_i^2 + 2a_B s_1$: $0 = \left(\sqrt{\frac{2ga}{3}}\right)^2 + 2(-g) s_1$ $0 = \frac{2ga}{3} - 2g s_1$ $s_1 = \frac{a}{3}$
  • Maximum height of B from the table = $2a + \frac{a}{3} = \frac{7a}{3}$.

• B falls back down and the string becomes taut:

  • A is resting on the table (height 0). The peg is at height 3a.
  • For the string to become taut, the length of the string on A's side (from peg to A) is 3a.
  • Therefore, the length of the string on B's side (from peg to B) must be $4a - 3a = a$.
  • This means B must fall until its height from the table is 'a'.
  • B starts falling from its maximum height $\frac{7a}{3}$ with initial velocity $u=0$.
  • Distance B needs to fall ($s_2$) = $\frac{7a}{3} - a = \frac{4a}{3}$.
  • Time taken for B to fall this distance ($t_2$): Using $s_2 = ut_2 + \frac{1}{2}gt_2^2$: $\frac{4a}{3} = 0 + \frac{1}{2} g t_2^2$ $t_2^2 = \frac{8a}{3g}$ $t_2 = \sqrt{\frac{8a}{3g}}$

• Total time A is resting on the table: This is the sum of the time B takes to go up and come back down until the string is taut. $T_{rest} = t_1 + t_2 = \sqrt{\frac{2a}{3g}} + \sqrt{\frac{8a}{3g}}$ $T_{rest} = \sqrt{\frac{2a}{3g}} + \sqrt{4 \cdot \frac{2a}{3g}}$ $T_{rest} = \sqrt{\frac{2a}{3g}} + 2\sqrt{\frac{2a}{3g}}$ $T_{rest} = 3\sqrt{\frac{2a}{3g}} = \sqrt{9 \cdot \frac{2a}{3g}} = \sqrt{\frac{18a}{3g}} = \sqrt{\frac{6a}{g}}$