Tiêu đề Low of motion (C.M. ) E.pdf ✅

Q.5 Consider the situation shown in figure. All pulley and strings are light. Acceleration of block m1, m2 and m3 are a1, a2 and a3 respectively. Assuming that all the blocks are undergoing pure translatory motion, match the following – m1 m2 2 m3 2 Column I Column II (A) m1 = 2 kg, (P) a1 = 0 m2 = 4 kg m3 = 6 kg (B) m1 = 4 kg, (Q) a2 is in upward m2 = 3.2 kg, direction m3 = 6 kg (C) m1 = m2 = 2 kg, (R) a3 = 0 m3 = 9 kg (D) m1 = 1 kg, (S) a3 is in downward m2 = 8 kg, direction m3 = 6 kg Ans. A  P, R ; B  Q, R ; C  P,Q,S ; D  R Q.6 Column I contains laws/principles and column II contain phenomena/things explained by them. Column-I Column-II (A) Newton's 1st law (P) Explanation of no effect of internal forces on the acceleration of system (B) Weak form of (Q) Definition of inertial Newton's 2nd law reference frame (C) Strong form of (R) Definition of force Newton's 2nd law (D) Newton's third law (S) Study of variable mass system Ans. A  Q ; B  R ; C  S ; D  P Q.7 Figure shows two blocks. A rough surface are identified as shown in figure. Blocks and man are initially at rest. The man started running with non zero acceleration on surface 'A' and jumps on surface 'C' and stays there stationary w.r.t. other block. Then - B D A C Column-I Column-II (A) Direction of contact force on 'A' when man is on 'A' (P) (B) Direction of contact force on surface B when man is on A (Q) (C) Direction of contact force on surface 'C' when man reaches at 'C' (R) (D) Direction of contact force on surface 'D' when man reaches at 'C' (S) Sol. A R ; B  P ; C  S ; D  Q Contact force on a surface is given by R N f      Where N  = normal force on the surface. f  = friction force on the surface. Q.8 Consider the situation shown in figure. All pulley and strings are light. Acceleration of block m1, m2 and m3 are a1, a2 and a3 respectively. Assuming that all the blocks are undergoing pure translatory motion, match the following – m1 m2 2 m3 2 Column I Column II (B) m1 = 2 kg, (P) a1 = 0 m2 = 4 kg m3 = 6 kg (B) m1 = 4 kg, (Q) a2 is in upward m2 = 3.2 kg, direction m3 = 6 kg (C) m1 = m2 = 2 kg, (R) a3 = 0 m3 = 9 kg (D) m1 = 1 kg, (S) a3 is in downward m2 = 8 kg, direction m3 = 6 kg Ans. A  P,R; B  Q,R ; C  P,Q,S ; D  R