Question

In the figure shown, upper block is given a velocity 6 m/s and very long plank, velocity 3m following quantities are to be matched when both attain same velocity.

• A. Work done by friction on 1 kg block in Joule
• B. Work done by friction on 2 kg plank in Joule
• C. Magnitude of change in momentum in N-s of 2kg plank
• D. Change in K.E. of system consisting of block and plank in joule
• E. Positive
• F. Negative
• G. 3
• H. 7
• I. 2

Answer 1

A → Q; B → P,S; C → T; D → Q,R

Answer 2

The problem involves two blocks, one on top of another, with initial velocities. We need to find the common final velocity when they move together, and then calculate work done by friction, change in momentum, and change in kinetic energy for various parts of the system.

1. Calculate the common final velocity ($v_f$): The system consists of the 1 kg block and the 2 kg plank. Since the surface below the plank is smooth, there are no external horizontal forces acting on the system. Therefore, the total linear momentum of the system is conserved.

Initial momentum of the system ($P_i$): $P_i = m_1 u_1 + m_2 u_2$ Given: $m_1 = 1 \text{ kg}$, $u_1 = 6 \text{ m/s}$ $m_2 = 2 \text{ kg}$, $u_2 = 3 \text{ m/s}$ $P_i = (1 \text{ kg})(6 \text{ m/s}) + (2 \text{ kg})(3 \text{ m/s}) = 6 + 6 = 12 \text{ kg m/s}$

Final momentum of the system ($P_f$): When both blocks attain the same velocity, they move together as a single unit with combined mass $(m_1 + m_2)$. $P_f = (m_1 + m_2) v_f = (1 \text{ kg} + 2 \text{ kg}) v_f = 3 v_f$

By conservation of momentum: $P_i = P_f$ $12 = 3 v_f$ $v_f = \frac{12}{3} = 4 \text{ m/s}$

2. Calculate quantities for Column-I:

(A) Work done by friction on 1 kg block in Joule: The 1 kg block slows down from $u_1 = 6 \text{ m/s}$ to $v_f = 4 \text{ m/s}$. The work done by friction on the 1 kg block is equal to its change in kinetic energy (Work-Energy Theorem). $W_{f1} = \Delta K_1 = \frac{1}{2} m_1 v_f^2 - \frac{1}{2} m_1 u_1^2$ $W_{f1} = \frac{1}{2} (1 \text{ kg}) (4 \text{ m/s})^2 - \frac{1}{2} (1 \text{ kg}) (6 \text{ m/s})^2$ $W_{f1} = \frac{1}{2} (16) - \frac{1}{2} (36) = 8 - 18 = -10 \text{ J}$ Since the work done is negative, (A) matches with (Q) Negative.

(B) Work done by friction on 2 kg plank in Joule: The 2 kg plank speeds up from $u_2 = 3 \text{ m/s}$ to $v_f = 4 \text{ m/s}$. The work done by friction on the 2 kg plank is equal to its change in kinetic energy. $W_{f2} = \Delta K_2 = \frac{1}{2} m_2 v_f^2 - \frac{1}{2} m_2 u_2^2$ $W_{f2} = \frac{1}{2} (2 \text{ kg}) (4 \text{ m/s})^2 - \frac{1}{2} (2 \text{ kg}) (3 \text{ m/s})^2$ $W_{f2} = (16) - (9) = 7 \text{ J}$ Since the work done is positive and its value is 7 J, (B) matches with (P) Positive and (S) 7.

(C) Magnitude of change in momentum in N-s of 2kg plank: Change in momentum of the 2 kg plank ($\Delta P_2$) is: $\Delta P_2 = m_2 v_f - m_2 u_2$ $\Delta P_2 = (2 \text{ kg})(4 \text{ m/s}) - (2 \text{ kg})(3 \text{ m/s}) = 8 - 6 = 2 \text{ N-s}$ The magnitude of change in momentum is 2 N-s. (C) matches with (T) 2.

(D) Change in K.E. of system consisting of block and plank in joule: Initial kinetic energy of the system ($K_{i,sys}$): $K_{i,sys} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$ $K_{i,sys} = \frac{1}{2} (1 \text{ kg}) (6 \text{ m/s})^2 + \frac{1}{2} (2 \text{ kg}) (3 \text{ m/s})^2$ $K_{i,sys} = \frac{1}{2} (36) + \frac{1}{2} (2)(9) = 18 + 9 = 27 \text{ J}$

Final kinetic energy of the system ($K_{f,sys}$): $K_{f,sys} = \frac{1}{2} (m_1 + m_2) v_f^2$ $K_{f,sys} = \frac{1}{2} (1 \text{ kg} + 2 \text{ kg}) (4 \text{ m/s})^2 = \frac{1}{2} (3)(16) = 24 \text{ J}$

Change in kinetic energy of the system ($\Delta K_{sys}$): $\Delta K_{sys} = K_{f,sys} - K_{i,sys} = 24 \text{ J} - 27 \text{ J} = -3 \text{ J}$ The change in kinetic energy is negative, and its magnitude is 3 J. (D) matches with (Q) Negative and (R) 3.