JEE Physics System Of Particles and Rotational Motion MCQs Set 38

Answer: (a) \( \frac{3(2+b)}{4(3+b)} \)

Explanation:
1. Mass \( M=\int_0^1a(1+bx^2)dx=a\left(1+\frac b3\right) \).
2. \( \int x\,dm=\int_0^1a(x+bx^3)dx=a\left(\frac12+\frac b4\right) \).
3. \( x_{cm}=\frac{\frac12+\frac b4}{1+\frac b3}=\frac{3(2+b)}{4(3+b)} \).

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. m
16. a
17. s
18. s
19.
20. o
21. f
22.
23. a
24.
25. c
26. o
27. n
28. t
29. i
30. n
31. u
32. o
33. u
34. s
35.
36. b
37. o
38. d
39. y
40.
41. i
42. s
43.
44. t
45. h
46. e
47.
48. i
49. n
50. t
51. e
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53. r
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56.
57. o
58. f
59.
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61.
62. d
63. m
64.
65. o
66. v
67. e
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69.
70. t
71. h
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73.
74. i
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78. g
79. r
80. a
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82.
83. o
84. f
85.
86. d
87. m
88. .
89.
90. C
91. o
92. m
93. b
94. i
95. n
96. e
97.
98. t
99. h
100. e
101.
102. f
103. r
104. a
105. c
106. t
107. i
108. o
109. n
110. s
111.
112. c
113. a
114. r
115. e
116. f
117. u
118. l
119. l
120. y
121. .

Answer: (b) \( \frac{2mu}{T} \)

Explanation:
1. Equal masses in an elastic head-on collision exchange velocities, so the target gets momentum \( mu \).
2. The impulse is the area of the triangle in the force-time graph: \( \frac12F_0T \).
3. \( \frac12F_0T=mu \), so \( F_0=\frac{2mu}{T} \).

Teacher's Note:
1. I
2. m
3. p
4. u
5. l
6. s
7. e
8.
9. e
10. q
11. u
12. a
13. l
14. s
15.
16. t
17. h
18. e
19.
20. a
21. r
22. e
23. a
24.
25. u
26. n
27. d
28. e
29. r
30.
31. t
32. h
33. e
34.
35. f
36. o
37. r
38. c
39. e
40. -
41. t
42. i
43. m
44. e
45.
46. g
47. r
48. a
49. p
50. h
51. .
52.
53. F
54. o
55. r
56.
57. a
58.
59. t
60. r
61. i
62. a
63. n
64. g
65. l
66. e
67.
68. t
69. h
70. a
71. t
72.
73. i
74. s
75.
76. h
77. a
78. l
79. f
80.
81. t
82. h
83. e
84.
85. b
86. a
87. s
88. e
89.
90. t
91. i
92. m
93. e
94. s
95.
96. t
97. h
98. e
99.
100. h
101. e
102. i
103. g
104. h
105. t
106. .

Answer: (c) \( u\sqrt{\sin^2\theta+e^2\cos^2\theta} \)

Explanation:
1. The floor is frictionless, so the horizontal component is unchanged: \( v_x=u\sin\theta \).
2. The vertical component becomes e times the original: \( v_y=eu\cos\theta \).
3. \( v=\sqrt{v_x^2+v_y^2}=u\sqrt{\sin^2\theta+e^2\cos^2\theta} \).

Teacher's Note:
1. R
2. e
3. s
4. o
5. l
6. v
7. e
8.
9. i
10. n
11. t
12. o
13.
14. c
15. o
16. m
17. p
18. o
19. n
20. e
21. n
22. t
23. s
24.
25. p
26. a
27. r
28. a
29. l
30. l
31. e
32. l
33.
34. a
35. n
36. d
37.
38. p
39. e
40. r
41. p
42. e
43. n
44. d
45. i
46. c
47. u
48. l
49. a
50. r
51.
52. t
53. o
54.
55. t
56. h
57. e
58.
59. s
60. u
61. r
62. f
63. a
64. c
65. e
66. .
67.
68. O
69. n
70. l
71. y
72.
73. t
74. h
75. e
76.
77. p
78. e
79. r
80. p
81. e
82. n
83. d
84. i
85. c
86. u
87. l
88. a
89. r
90.
91. c
92. o
93. m
94. p
95. o
96. n
97. e
98. n
99. t
100.
101. i
102. s
103.
104. m
105. u
106. l
107. t
108. i
109. p
110. l
111. i
112. e
113. d
114.
115. b
116. y
117.
118. e
119. .

Answer: (d) v in the direction \( \overrightarrow{BO} \)

Explanation:
1. Initially the three momenta have zero resultant, so the total momentum afterwards is also zero.
2. Retracing means B now moves from O towards B with velocity v along \( \overrightarrow{OB} \). A is at rest.
3. So \( m\vec v_C=-m\vec v_B \): C moves with speed v along \( \overrightarrow{BO} \).

Teacher's Note:
1. T
2. h
3. e
4.
5. t
6. o
7. t
8. a
9. l
10.
11. m
12. o
13. m
14. e
15. n
16. t
17. u
18. m
19.
20. i
21. s
22.
23. z
24. e
25. r
26. o
27.
28. b
29. e
30. f
31. o
32. r
33. e
34.
35. t
36. h
37. e
38.
39. c
40. o
41. l
42. l
43. i
44. s
45. i
46. o
47. n
48.
49. a
50. n
51. d
52.
53. m
54. u
55. s
56. t
57.
58. b
59. e
60.
61. z
62. e
63. r
64. o
65.
66. a
67. f
68. t
69. e
70. r
71.
72. i
73. t
74. .
75.
76. W
77. h
78. a
79. t
80. e
81. v
82. e
83. r
84.
85. B
86.
87. a
88. n
89. d
90.
91. A
92.
93. c
94. a
95. r
96. r
97. y
98. ,
99.
100. C
101.
102. m
103. u
104. s
105. t
106.
107. c
108. a
109. n
110. c
111. e
112. l
113. .

Answer: (a) The coefficient of restitution is 1

Explanation:
1. The angle in a semicircle is a right angle, so \( \angle PQS=90^\circ \) and QS makes \( 45^\circ \) with PS too; Q is at the top of the circle and the normal at Q is vertical.
2. The incident ray PQ and the reflected ray QS make \( 45^\circ \) with the normal on either side: the angle of reflection equals the angle of incidence and the speed is unchanged.
3. For the normal component to be reversed with equal magnitude, \( e=1 \), and the collision is perfectly elastic.

Teacher's Note:
1. A
2.
3. s
4. m
5. o
6. o
7. t
8. h
9.
10. w
11. a
12. l
13. l
14.
15. c
16. h
17. a
18. n
19. g
20. e
21. s
22.
23. o
24. n
25. l
26. y
27.
28. t
29. h
30. e
31.
32. n
33. o
34. r
35. m
36. a
37. l
38.
39. c
40. o
41. m
42. p
43. o
44. n
45. e
46. n
47. t
48. .
49.
50. E
51. q
52. u
53. a
54. l
55.
56. a
57. n
58. g
59. l
60. e
61. s
62.
63. o
64. f
65.
66. i
67. n
68. c
69. i
70. d
71. e
72. n
73. c
74. e
75.
76. a
77. n
78. d
79.
80. r
81. e
82. f
83. l
84. e
85. c
86. t
87. i
88. o
89. n
90.
91. i
92. m
93. p
94. l
95. y
96.
97. t
98. h
99. e
100.
101. n
102. o
103. r
104. m
105. a
106. l
107.
108. c
109. o
110. m
111. p
112. o
113. n
114. e
115. n
116. t
117.
118. i
119. s
120.
121. r
122. e
123. v
124. e
125. r
126. s
127. e
128. d
129.
130. w
131. i
132. t
133. h
134. o
135. u
136. t
137.
138. l
139. o
140. s
141. s
142. .

Answer: (b) \( y=2x \)

Explanation:
1. Velocity of the centre of mass \( =\frac{m(v\hat i)+2m(v\hat j)}{3m}=\frac v3(\hat i+2\hat j) \).
2. The centre of mass starts at the origin and moves along the direction \( \hat i+2\hat j \).
3. So its path is \( y=2x \).

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
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15. m
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17. s
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19.
20. m
21. o
22. v
23. e
24. s
25.
26. i
27. n
28.
29. t
30. h
31. e
32.
33. d
34. i
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36. e
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40. o
41. n
42.
43. o
44. f
45.
46. t
47. h
48. e
49.
50. t
51. o
52. t
53. a
54. l
55.
56. m
57. o
58. m
59. e
60. n
61. t
62. u
63. m
64. .
65.
66. S
67. t
68. a
69. r
70. t
71. i
72. n
73. g
74.
75. f
76. r
77. o
78. m
79.
80. t
81. h
82. e
83.
84. o
85. r
86. i
87. g
88. i
89. n
90. ,
91.
92. t
93. h
94. e
95.
96. l
97. i
98. n
99. e
100.
101. f
102. o
103. l
104. l
105. o
106. w
107. s
108.
109. t
110. h
111. e
112.
113. v
114. e
115. l
116. o
117. c
118. i
119. t
120. y
121.
122. r
123. a
124. t
125. i
126. o
127. .

Answer: (c) \( -\hat i+2\hat j \)

Explanation:
1. The wall is parallel to \( \hat j \), so the normal to the wall is along \( \hat i \).
2. The \( \hat i \) component reverses and is multiplied by e: \( 2\to-2\times\frac12=-1 \).
3. The \( \hat j \) component is unchanged (smooth wall): 2. So the velocity is \( -\hat i+2\hat j \).

Teacher's Note:
1. O
2. n
3. l
4. y
5.
6. t
7. h
8. e
9.
10. c
11. o
12. m
13. p
14. o
15. n
16. e
17. n
18. t
19.
20. p
21. e
22. r
23. p
24. e
25. n
26. d
27. i
28. c
29. u
30. l
31. a
32. r
33.
34. t
35. o
36.
37. t
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39. e
40.
41. w
42. a
43. l
44. l
45.
46. c
47. h
48. a
49. n
50. g
51. e
52. s
53. .
54.
55. T
56. h
57. e
58.
59. p
60. a
61. r
62. a
63. l
64. l
65. e
66. l
67.
68. c
69. o
70. m
71. p
72. o
73. n
74. e
75. n
76. t
77.
78. s
79. t
80. a
81. y
82. s
83.
84. t
85. h
86. e
87.
88. s
89. a
90. m
91. e
92.
93. o
94. n
95.
96. a
97.
98. s
99. m
100. o
101. o
102. t
103. h
104.
105. w
106. a
107. l
108. l
109. .

Answer: (d) \( \frac13 \)

Explanation:
1. Let the first ball (mass \( m_1 \)) have speed u and the second (mass \( m_2 \)) end with speed \( v_2 \). Momentum: \( m_1u=m_2v_2 \). Restitution: \( e=\frac{v_2}{u} \).
2. Final kinetic energy \( =\frac12m_2v_2^2=\frac12(m_1u)v_2 \). Ratio to initial \( =\frac{m_1uv_2}{m_1u^2}=\frac{v_2}{u}=e \).
3. One-third of the energy remains, so \( e=\frac13 \).

Teacher's Note:
1. C
2. o
3. m
4. b
5. i
6. n
7. e
8.
9. m
10. o
11. m
12. e
13. n
14. t
15. u
16. m
17.
18. c
19. o
20. n
21. s
22. e
23. r
24. v
25. a
26. t
27. i
28. o
29. n
30.
31. w
32. i
33. t
34. h
35.
36. t
37. h
38. e
39.
40. d
41. e
42. f
43. i
44. n
45. i
46. t
47. i
48. o
49. n
50.
51. o
52. f
53.
54. e
55. .
56.
57. T
58. h
59. e
60.
61. f
62. i
63. n
64. a
65. l
66. -
67. t
68. o
69. -
70. i
71. n
72. i
73. t
74. i
75. a
76. l
77.
78. e
79. n
80. e
81. r
82. g
83. y
84.
85. r
86. a
87. t
88. i
89. o
90.
91. t
92. h
93. e
94. n
95.
96. t
97. u
98. r
99. n
100. s
101.
102. o
103. u
104. t
105.
106. t
107. o
108.
109. e
110. q
111. u
112. a
113. l
114.
115. e
116. .

Answer: (a) 0

Explanation:
1. Speed before the strike \( =\sqrt{60} \) m/s, along the \( 60^\circ \) incline. The angle between this direction and the second incline is \( 30^\circ \).
2. Component along the second incline \( =\sqrt{45} \) (down the \( 30^\circ \) slope); perpendicular component \( \sqrt{15} \) reverses to point away from the incline.
3. Vertical component \( =-\sqrt{45}\sin30^\circ+\sqrt{15}\cos30^\circ=-\frac{\sqrt{45}}{2}+\frac{\sqrt{45}}{2}=0 \).

Teacher's Note:
1. S
2. p
3. l
4. i
5. t
6.
7. t
8. h
9. e
10.
11. v
12. e
13. l
14. o
15. c
16. i
17. t
18. y
19.
20. a
21. l
22. o
23. n
24. g
25.
26. a
27. n
28. d
29.
30. p
31. e
32. r
33. p
34. e
35. n
36. d
37. i
38. c
39. u
40. l
41. a
42. r
43.
44. t
45. o
46.
47. t
48. h
49. e
50.
51. s
52. e
53. c
54. o
55. n
56. d
57.
58. i
59. n
60. c
61. l
62. i
63. n
64. e
65. ,
66.
67. r
68. e
69. v
70. e
71. r
72. s
73. e
74.
75. t
76. h
77. e
78.
79. p
80. e
81. r
82. p
83. e
84. n
85. d
86. i
87. c
88. u
89. l
90. a
91. r
92.
93. p
94. a
95. r
96. t
97. ,
98.
99. a
100. n
101. d
102.
103. t
104. h
105. e
106. n
107.
108. r
109. e
110. s
111. o
112. l
113. v
114. e
115.
116. b
117. o
118. t
119. h
120.
121. v
122. e
123. r
124. t
125. i
126. c
127. a
128. l
129. l
130. y
131. .

Answer: (b) \( \left(\frac mM\right)v\cos\alpha \)

Explanation:
1. Horizontal momentum is conserved. The car is to be at rest after firing, so the shell's speed relative to the ground equals its speed relative to the barrel.
2. Horizontal component of the shell's velocity \( =v\cos\alpha \).
3. \( MV=m\,v\cos\alpha \), so \( V=\frac mMv\cos\alpha \).

Teacher's Note:
1. O
2. n
3. l
4. y
5.
6. t
7. h
8. e
9.
10. h
11. o
12. r
13. i
14. z
15. o
16. n
17. t
18. a
19. l
20.
21. m
22. o
23. m
24. e
25. n
26. t
27. u
28. m
29.
30. i
31. s
32.
33. c
34. o
35. n
36. s
37. e
38. r
39. v
40. e
41. d
42.
43. b
44. e
45. c
46. a
47. u
48. s
49. e
50.
51. t
52. h
53. e
54.
55. g
56. r
57. o
58. u
59. n
60. d
61.
62. c
63. a
64. n
65.
66. s
67. u
68. p
69. p
70. l
71. y
72.
73. v
74. e
75. r
76. t
77. i
78. c
79. a
80. l
81.
82. f
83. o
84. r
85. c
86. e
87. s
88. .
89.
90. I
91. f
92.
93. t
94. h
95. e
96.
97. c
98. a
99. r
100.
101. s
102. t
103. o
104. p
105. s
106. ,
107.
108. t
109. h
110. e
111.
112. s
113. h
114. e
115. l
116. l
117. '
118. s
119.
120. g
121. r
122. o
123. u
124. n
125. d
126.
127. s
128. p
129. e
130. e
131. d
132.
133. e
134. q
135. u
136. a
137. l
138. s
139.
140. i
141. t
142. s
143.
144. s
145. p
146. e
147. e
148. d
149.
150. r
151. e
152. l
153. a
154. t
155. i
156. v
157. e
158.
159. t
160. o
161.
162. t
163. h
164. e
165.
166. b
167. a
168. r
169. r
170. e
171. l
172. .

Answer: (c) I and IV only

Explanation:
1. At the top the shell moves horizontally with speed u. Part A retraces its path, so its velocity is \( -u \).
2. Momentum conservation: \( 2mu=m(-u)+mv_B \), so \( v_B=3u \).
3. B has three times the speed, so nine times the kinetic energy (IV true). Its horizontal travel after the top is 3 times \( \frac R2 \), so it lands \( \frac R2+\frac{3R}{2}=2R \) from the cannon, i.e. a distance R beyond the target (I true; II and III false).

Teacher's Note:
1. A
2. t
3.
4. t
5. h
6. e
7.
8. t
9. o
10. p
11.
12. o
13. f
14.
15. t
16. h
17. e
18.
19. p
20. a
21. t
22. h
23.
24. o
25. n
26. l
27. y
28.
29. t
30. h
31. e
32.
33. h
34. o
35. r
36. i
37. z
38. o
39. n
40. t
41. a
42. l
43.
44. m
45. o
46. m
47. e
48. n
49. t
50. u
51. m
52.
53. e
54. x
55. i
56. s
57. t
58. s
59. .
60.
61. T
62. h
63. e
64.
65. t
66. i
67. m
68. e
69.
70. o
71. f
72.
73. f
74. a
75. l
76. l
77.
78. f
79. o
80. r
81.
82. b
83. o
84. t
85. h
86.
87. p
88. a
89. r
90. t
91. s
92.
93. i
94. s
95.
96. t
97. h
98. e
99.
100. s
101. a
102. m
103. e
104. ,
105.
106. s
107. o
108.
109. d
110. i
111. s
112. t
113. a
114. n
115. c
116. e
117. s
118.
119. s
120. c
121. a
122. l
123. e
124.
125. w
126. i
127. t
128. h
129.
130. s
131. p
132. e
133. e
134. d
135. .

Answer: (d) \( 120^\circ \)

Explanation:
1. Let each have mass m and speed v with angle \( \theta \) between the velocities. The vector sum has magnitude \( 2mv\cos\frac\theta2 \).
2. Common velocity \( =\frac{2mv\cos(\theta/2)}{2m}=v\cos\frac\theta2=\frac v2 \).
3. \( \cos\frac\theta2=\frac12 \), so \( \theta=120^\circ \).

Teacher's Note:
1. A
2. d
3. d
4.
5. t
6. h
7. e
8.
9. t
10. w
11. o
12.
13. e
14. q
15. u
16. a
17. l
18.
19. m
20. o
21. m
22. e
23. n
24. t
25. a
26.
27. a
28. s
29.
30. v
31. e
32. c
33. t
34. o
35. r
36. s
37. .
38.
39. T
40. h
41. e
42.
43. r
44. e
45. s
46. u
47. l
48. t
49. a
50. n
51. t
52.
53. b
54. i
55. s
56. e
57. c
58. t
59. s
60.
61. t
62. h
63. e
64.
65. a
66. n
67. g
68. l
69. e
70.
71. b
72. e
73. t
74. w
75. e
76. e
77. n
78.
79. t
80. h
81. e
82. m
83. .

Answer: (a) \( \sqrt{2gh\left(1+\frac1\eta\right)} \)

Explanation:
1. At the minimum speed the block just reaches the top, moving with the wedge: common horizontal velocity V. Momentum: \( mu=(m+\eta m)V \).
2. Energy: \( \frac12mu^2=\frac12(m+\eta m)V^2+mgh \).
3. \( u^2\left(1-\frac{1}{1+\eta}\right)=2gh\Rightarrow u^2\frac{\eta}{1+\eta}=2gh \), so \( u=\sqrt{2gh\left(1+\frac1\eta\right)} \).

Teacher's Note:
1. A
2. t
3.
4. t
5. h
6. e
7.
8. h
9. i
10. g
11. h
12. e
13. s
14. t
15.
16. p
17. o
18. i
19. n
20. t
21.
22. t
23. h
24. e
25.
26. b
27. l
28. o
29. c
30. k
31.
32. h
33. a
34. s
35.
36. n
37. o
38.
39. v
40. e
41. l
42. o
43. c
44. i
45. t
46. y
47.
48. r
49. e
50. l
51. a
52. t
53. i
54. v
55. e
56.
57. t
58. o
59.
60. t
61. h
62. e
63.
64. w
65. e
66. d
67. g
68. e
69. .
70.
71. U
72. s
73. e
74.
75. h
76. o
77. r
78. i
79. z
80. o
81. n
82. t
83. a
84. l
85.
86. m
87. o
88. m
89. e
90. n
91. t
92. u
93. m
94.
95. c
96. o
97. n
98. s
99. e
100. r
101. v
102. a
103. t
104. i
105. o
106. n
107.
108. t
109. o
110. g
111. e
112. t
113. h
114. e
115. r
116.
117. w
118. i
119. t
120. h
121.
122. e
123. n
124. e
125. r
126. g
127. y
128.
129. c
130. o
131. n
132. s
133. e
134. r
135. v
136. a
137. t
138. i
139. o
140. n
141. .

Answer: (b) \( \frac v2 \)

Explanation:
1. Take the direction of the first particle as positive: velocities are 2v and \( -v \).
2. \( v_{cm}=\frac{m(2v)+m(-v)}{2m} \).
3. \( v_{cm}=\frac v2 \).

Teacher's Note:
1. V
2. e
3. l
4. o
5. c
6. i
7. t
8. i
9. e
10. s
11.
12. a
13. r
14. e
15.
16. v
17. e
18. c
19. t
20. o
21. r
22. s
23. .
24.
25. T
26. h
27. e
28.
29. p
30. a
31. r
32. t
33. i
34. c
35. l
36. e
37.
38. m
39. o
40. v
41. i
42. n
43. g
44.
45. t
46. h
47. e
48.
49. o
50. t
51. h
52. e
53. r
54.
55. w
56. a
57. y
58.
59. c
60. o
61. n
62. t
63. r
64. i
65. b
66. u
67. t
68. e
69. s
70.
71. a
72.
73. n
74. e
75. g
76. a
77. t
78. i
79. v
80. e
81.
82. s
83. i
84. g
85. n
86. .

Answer: (c) I does not imply II, but II implies I

Explanation:
1. If the kinetic energy is zero, every particle must be at rest, so the momentum is zero: II implies I.
2. If the total momentum is zero, the particles may still move in opposite directions (as after an explosion), giving a non-zero kinetic energy.
3. So I does not imply II.

Teacher's Note:
1. M
2. o
3. m
4. e
5. n
6. t
7. u
8. m
9.
10. i
11. s
12.
13. a
14.
15. v
16. e
17. c
18. t
19. o
20. r
21.
22. s
23. u
24. m
25.
26. t
27. h
28. a
29. t
30.
31. c
32. a
33. n
34.
35. c
36. a
37. n
38. c
39. e
40. l
41. ,
42.
43. b
44. u
45. t
46.
47. k
48. i
49. n
50. e
51. t
52. i
53. c
54.
55. e
56. n
57. e
58. r
59. g
60. y
61.
62. i
63. s
64.
65. a
66.
67. s
68. u
69. m
70.
71. o
72. f
73.
74. p
75. o
76. s
77. i
78. t
79. i
80. v
81. e
82.
83. t
84. e
85. r
86. m
87. s
88.
89. a
90. n
91. d
92.
93. c
94. a
95. n
96. n
97. o
98. t
99. .

Answer: (d) \( \frac{2v}{\sqrt3} \)

Explanation:
1. Momentum along the original direction: \( mv=mv_x \), so \( v_x=v \).
2. Perpendicular direction: \( 0=m\frac{v}{\sqrt3}-mv_y \), so \( v_y=\frac{v}{\sqrt3} \) (in the opposite sense to the first mass).
3. Speed \( =\sqrt{v^2+\frac{v^2}{3}}=\frac{2v}{\sqrt3} \).

Teacher's Note:
1. A
2. p
3. p
4. l
5. y
6.
7. c
8. o
9. n
10. s
11. e
12. r
13. v
14. a
15. t
16. i
17. o
18. n
19.
20. o
21. f
22.
23. m
24. o
25. m
26. e
27. n
28. t
29. u
30. m
31.
32. s
33. e
34. p
35. a
36. r
37. a
38. t
39. e
40. l
41. y
42.
43. a
44. l
45. o
46. n
47. g
48.
49. a
50. n
51. d
52.
53. p
54. e
55. r
56. p
57. e
58. n
59. d
60. i
61. c
62. u
63. l
64. a
65. r
66.
67. t
68. o
69.
70. t
71. h
72. e
73.
74. o
75. r
76. i
77. g
78. i
79. n
80. a
81. l
82.
83. d
84. i
85. r
86. e
87. c
88. t
89. i
90. o
91. n
92. ,
93.
94. t
95. h
96. e
97. n
98.
99. c
100. o
101. m
102. b
103. i
104. n
105. e
106.
107. t
108. h
109. e
110.
111. c
112. o
113. m
114. p
115. o
116. n
117. e
118. n
119. t
120. s
121. .

Answer: (a) 288 J

Explanation:
1. Momentum conservation: \( 4v=12\times4 \), so \( v=12 \) m/s.
2. \( K=\frac12(4)(12)^2 \).
3. \( K=288 \) J.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. m
7. a
8. l
9. l
10. e
11. r
12.
13. p
14. i
15. e
16. c
17. e
18.
19. m
20. o
21. v
22. e
23. s
24.
25. f
26. a
27. s
28. t
29. e
30. r
31. .
32.
33. F
34. i
35. n
36. d
37.
38. i
39. t
40. s
41.
42. s
43. p
44. e
45. e
46. d
47.
48. f
49. i
50. r
51. s
52. t
53. ,
54.
55. t
56. h
57. e
58. n
59.
60. i
61. t
62. s
63.
64. k
65. i
66. n
67. e
68. t
69. i
70. c
71.
72. e
73. n
74. e
75. r
76. g
77. y
78. .

Answer: (b) \( \frac13 \)

Explanation:
1. Masses are proportional to areas: whole disc \( 4\pi R^2\sigma \), removed disc \( \pi R^2\sigma \), remaining \( 3\pi R^2\sigma \).
2. The removed disc's centre is at distance R from the centre of the big disc.
3. \( x_{cm}=\frac{4(0)-1(R)}{3}=-\frac R3 \): the distance is \( \frac R3 \), so \( \alpha=\frac13 \).

Teacher's Note:
1. T
2. r
3. e
4. a
5. t
6.
7. t
8. h
9. e
10.
11. r
12. e
13. m
14. o
15. v
16. e
17. d
18.
19. p
20. a
21. r
22. t
23.
24. a
25. s
26.
27. a
28.
29. n
30. e
31. g
32. a
33. t
34. i
35. v
36. e
37.
38. m
39. a
40. s
41. s
42.
43. a
44. t
45.
46. i
47. t
48. s
49.
50. o
51. w
52. n
53.
54. c
55. e
56. n
57. t
58. r
59. e
60. .
61.
62. T
63. h
64. e
65.
66. a
67. r
68. e
69. a
70.
71. r
72. a
73. t
74. i
75. o
76.
77. g
78. i
79. v
80. e
81. s
82.
83. t
84. h
85. e
86.
87. m
88. a
89. s
90. s
91.
92. r
93. a
94. t
95. i
96. o
97.
98. f
99. o
100. r
101.
102. a
103.
104. u
105. n
106. i
107. f
108. o
109. r
110. m
111.
112. d
113. i
114. s
115. c
116. .

Answer: (c) 56%

Explanation:
1. Momentum: \( p_x=2mv \), \( p_y=2mv \), total mass \( 3m \), so \( V=\frac{2\sqrt2mv}{3m}=\frac{2\sqrt2}{3}v \).
2. Initial energy \( =\frac12m(4v^2)+\frac12(2m)v^2=3mv^2 \). Final energy \( =\frac12(3m)\frac{8v^2}{9}=\frac43mv^2 \).
3. Fractional loss \( =1-\frac49=\frac59\approx56\% \).

Teacher's Note:
1. C
2. o
3. m
4. b
5. i
6. n
7. e
8.
9. t
10. h
11. e
12.
13. t
14. w
15. o
16.
17. p
18. e
19. r
20. p
21. e
22. n
23. d
24. i
25. c
26. u
27. l
28. a
29. r
30.
31. m
32. o
33. m
34. e
35. n
36. t
37. a
38.
39. a
40. s
41.
42. v
43. e
44. c
45. t
46. o
47. r
48. s
49. .
50.
51. C
52. o
53. m
54. p
55. a
56. r
57. e
58.
59. t
60. h
61. e
62.
63. k
64. i
65. n
66. e
67. t
68. i
69. c
70.
71. e
72. n
73. e
74. r
75. g
76. i
77. e
78. s
79.
80. b
81. e
82. f
83. o
84. r
85. e
86.
87. a
88. n
89. d
90.
91. a
92. f
93. t
94. e
95. r
96. .

Answer: (d) 4

Explanation:
1. In the frame of the heavy wall the ball approaches at \( 2+1=3 \) m/s.
2. An elastic collision with a heavy wall reverses this relative velocity, so the ball leaves at 3 m/s relative to the wall.
3. The wall moves at 1 m/s towards the ball's original side, so relative to the ground the ball moves at \( 3+1=4 \) m/s.

Teacher's Note:
1. S
2. w
3. i
4. t
5. c
6. h
7.
8. t
9. o
10.
11. t
12. h
13. e
14.
15. f
16. r
17. a
18. m
19. e
20.
21. o
22. f
23.
24. t
25. h
26. e
27.
28. w
29. a
30. l
31. l
32. ,
33.
34. r
35. e
36. v
37. e
38. r
39. s
40. e
41.
42. t
43. h
44. e
45.
46. r
47. e
48. l
49. a
50. t
51. i
52. v
53. e
54.
55. v
56. e
57. l
58. o
59. c
60. i
61. t
62. y
63.
64. a
65. n
66. d
67.
68. t
69. h
70. e
71. n
72.
73. s
74. w
75. i
76. t
77. c
78. h
79.
80. b
81. a
82. c
83. k
84. .

Answer: (a) 4

Explanation:
1. The sand starts with no horizontal velocity and must be brought to 0.8 m/s.
2. \( F=v\frac{dm}{dt} \).
3. \( F=0.8\times5=4 \) N.

Teacher's Note:
1. F
2. o
3. r
4. c
5. e
6.
7. o
8. n
9.
10. t
11. h
12. e
13.
14. b
15. e
16. l
17. t
18.
19. e
20. q
21. u
22. a
23. l
24. s
25.
26. t
27. h
28. e
29.
30. r
31. a
32. t
33. e
34.
35. a
36. t
37.
38. w
39. h
40. i
41. c
42. h
43.
44. h
45. o
46. r
47. i
48. z
49. o
50. n
51. t
52. a
53. l
54.
55. m
56. o
57. m
58. e
59. n
60. t
61. u
62. m
63.
64. i
65. s
66.
67. g
68. i
69. v
70. e
71. n
72.
73. t
74. o
75.
76. t
77. h
78. e
79.
80. s
81. a
82. n
83. d
84. .

Answer: (b) 100

Explanation:
1. Heights: \( y_1=50t-5t^2 \) and \( y_2=40+30t-5t^2 \).
2. Centre of mass height \( y_{cm}=\frac{y_1+y_2}{2}=20+40t-5t^2 \).
3. Maximum when \( 40-10t=0 \), \( t=4 \) s: \( y_{cm}=20+160-80=100 \) m.

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. m
16. a
17. s
18. s
19.
20. m
21. o
22. v
23. e
24. s
25.
26. l
27. i
28. k
29. e
30.
31. a
32.
33. s
34. i
35. n
36. g
37. l
38. e
39.
40. p
41. a
42. r
43. t
44. i
45. c
46. l
47. e
48.
49. w
50. i
51. t
52. h
53.
54. t
55. h
56. e
57.
58. a
59. v
60. e
61. r
62. a
63. g
64. e
65.
66. i
67. n
68. i
69. t
70. i
71. a
72. l
73.
74. h
75. e
76. i
77. g
78. h
79. t
80.
81. a
82. n
83. d
84.
85. v
86. e
87. l
88. o
89. c
90. i
91. t
92. y
93.
94. u
95. n
96. d
97. e
98. r
99.
100. g
101. r
102. a
103. v
104. i
105. t
106. y
107. .

Answer: (c) 5

Explanation:
1. At maximum compression the blocks move together: \( V=\frac{2\times0.15}{5}=0.06 \) m/s.
2. Energy: \( \frac12kx^2=\frac12(2)(0.15)^2-\frac12(5)(0.06)^2=0.0225-0.009=0.0135 \) J.
3. \( x^2=\frac{0.027}{10.8}=0.0025 \), so \( x=0.05 \) m \( =5 \) cm.

Teacher's Note:
1. A
2. t
3.
4. t
5. h
6. e
7.
8. i
9. n
10. s
11. t
12. a
13. n
14. t
15.
16. o
17. f
18.
19. m
20. a
21. x
22. i
23. m
24. u
25. m
26.
27. c
28. o
29. m
30. p
31. r
32. e
33. s
34. s
35. i
36. o
37. n
38.
39. t
40. h
41. e
42.
43. r
44. e
45. l
46. a
47. t
48. i
49. v
50. e
51.
52. v
53. e
54. l
55. o
56. c
57. i
58. t
59. y
60.
61. i
62. s
63.
64. z
65. e
66. r
67. o
68. .
69.
70. T
71. h
72. e
73.
74. l
75. o
76. s
77. t
78.
79. k
80. i
81. n
82. e
83. t
84. i
85. c
86.
87. e
88. n
89. e
90. r
91. g
92. y
93.
94. i
95. s
96.
97. s
98. t
99. o
100. r
101. e
102. d
103.
104. i
105. n
106.
107. t
108. h
109. e
110.
111. s
112. p
113. r
114. i
115. n
116. g
117. .

Answer: (d) \( 5.1\ \text{m/s}^2 \) downward

Explanation:
1. Acceleration of each mass: \( a=\frac{6-1}{6+1}g=\frac{50}{7} \) m/s² (6 kg down, 1 kg up).
2. \( a_{cm}=\frac{6a-1a}{7}=\frac{5a}{7} \).
3. \( a_{cm}=\frac{5}{7}\times\frac{50}{7}=\frac{250}{49}\approx5.1 \) m/s² downward.

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. m
16. a
17. s
18. s
19.
20. a
21. c
22. c
23. e
24. l
25. e
26. r
27. a
28. t
29. i
30. o
31. n
32.
33. i
34. s
35.
36. t
37. h
38. e
39.
40. m
41. a
42. s
43. s
44. -
45. w
46. e
47. i
48. g
49. h
50. t
51. e
52. d
53.
54. a
55. v
56. e
57. r
58. a
59. g
60. e
61.
62. o
63. f
64.
65. t
66. h
67. e
68.
69. t
70. w
71. o
72.
73. a
74. c
75. c
76. e
77. l
78. e
79. r
80. a
81. t
82. i
83. o
84. n
85. s
86. ,
87.
88. t
89. a
90. k
91. i
92. n
93. g
94.
95. t
96. h
97. e
98.
99. d
100. i
101. r
102. e
103. c
104. t
105. i
106. o
107. n
108.
109. o
110. f
111.
112. m
113. o
114. t
115. i
116. o
117. n
118.
119. i
120. n
121. t
122. o
123.
124. a
125. c
126. c
127. o
128. u
129. n
130. t
131. .