Centre of Mass and its Motion Practice Questions - DPP for NEET

 Page 1


1. (c) m
1
 = 1, m
2
 = 35.5
1
, 
12
ˆ
0, 1.27 r ri
rr
==
11 22
12
mr mr
r
mm
rr
r ´
=
+
 Þ
35.5 1.27
ˆ
1 35.5
ri
´
=
+
r
35.5
ˆˆ
1.27 1.24
36.5
r ii = ´=
r
1
m 2
m
1.27Å
y
H
Cl
x
2. (d)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 3 3 2 12
2.4 m/s
235
´+´
= ==
+
3. (c) m
1
 =  12, m
2
 = 16
12
ˆˆ ˆˆ
0 0, 1.10 r i jr ij
r
= + =+
11 22
1
12
mr mr
r
mm
+
=
+
rr
r
C O
m
2
m
1
x
y
1
16 1.1
ˆˆ
0.63
28
r ii
´
==
r
i.e. 0.63Å from carbon atom.
4. (a)
1 1 2 2 33
1 23
cm
mv m v mv
v
m mm
++
=
++
r rr
r
= 
ˆ ˆˆ
20 10 30 10 50 10
100
i jk ´ +´ +´
\ 
ˆ ˆˆ
2 35
cm
v i jk =++
5. (c) We can assume that three particles of equal mass m are
placed at the corners of triangle.
12
ˆˆ ˆˆ
r 0i 0j,r bi 0j = + =+
u r ur
and 
3
ˆˆ
r 0i hj =+
ur
\ 
11 2 2 33
cm
1 23
m r m r mr
r
m mm
++
=
++
u r u r ur
uuu r
 
x
y
(b,0)
(0,0)
(0,h)
= 
bh
ˆˆ
ij
33
+
i.e. coordinates of centre of mass is 
bh
,
33
æö
ç÷
èø
6. (b)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
       
224 10
7.3m/s
24
´ +´
==
+
7. (b) Let m
1
 = m, m
2
 = 2m, m
3
 = 3m, m
4
 = 4m
a
2m
m
3m
4m
x
y
60°
a cos 60°
a sin 60°
1
ˆˆ
00 r ij
r
=+
2
3
ˆ ˆˆ
r cos60 sin 60
22
aa
a ia j ij =+=+
r
3
33
ˆ ˆ ˆˆ
r ( cos 60) sin 60
22
a
aa ia j aij
r
= + + =+
4
ˆˆ
r0 aij =+
r
by substituting above value in the following formula
11 2 2 3 3 44
1 2 34
3
ˆ
r 0.95
4
mr m r m r mr
ai aj
m m mm
+ ++
= =+
+ ++
r r rr
r
So the location of centre of mass 
3
0.95 ,
4
aa
éù
êú
ëû
8. (d)
9. (d) m
1
 = 2kg, m
2
 = 4kg, 
12
2/ , 10/ v msv ms = =-
rr
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 20 4 10
0/
24
ms
´ -´
==
+
10. (a) As initially both the particles were at rest therefore
velocity of centre of mass was zero and there is no
external force on the system so speed of centre of mass
remains constant i.e. it should be equal to zero.
Page 2


1. (c) m
1
 = 1, m
2
 = 35.5
1
, 
12
ˆ
0, 1.27 r ri
rr
==
11 22
12
mr mr
r
mm
rr
r ´
=
+
 Þ
35.5 1.27
ˆ
1 35.5
ri
´
=
+
r
35.5
ˆˆ
1.27 1.24
36.5
r ii = ´=
r
1
m 2
m
1.27Å
y
H
Cl
x
2. (d)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 3 3 2 12
2.4 m/s
235
´+´
= ==
+
3. (c) m
1
 =  12, m
2
 = 16
12
ˆˆ ˆˆ
0 0, 1.10 r i jr ij
r
= + =+
11 22
1
12
mr mr
r
mm
+
=
+
rr
r
C O
m
2
m
1
x
y
1
16 1.1
ˆˆ
0.63
28
r ii
´
==
r
i.e. 0.63Å from carbon atom.
4. (a)
1 1 2 2 33
1 23
cm
mv m v mv
v
m mm
++
=
++
r rr
r
= 
ˆ ˆˆ
20 10 30 10 50 10
100
i jk ´ +´ +´
\ 
ˆ ˆˆ
2 35
cm
v i jk =++
5. (c) We can assume that three particles of equal mass m are
placed at the corners of triangle.
12
ˆˆ ˆˆ
r 0i 0j,r bi 0j = + =+
u r ur
and 
3
ˆˆ
r 0i hj =+
ur
\ 
11 2 2 33
cm
1 23
m r m r mr
r
m mm
++
=
++
u r u r ur
uuu r
 
x
y
(b,0)
(0,0)
(0,h)
= 
bh
ˆˆ
ij
33
+
i.e. coordinates of centre of mass is 
bh
,
33
æö
ç÷
èø
6. (b)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
       
224 10
7.3m/s
24
´ +´
==
+
7. (b) Let m
1
 = m, m
2
 = 2m, m
3
 = 3m, m
4
 = 4m
a
2m
m
3m
4m
x
y
60°
a cos 60°
a sin 60°
1
ˆˆ
00 r ij
r
=+
2
3
ˆ ˆˆ
r cos60 sin 60
22
aa
a ia j ij =+=+
r
3
33
ˆ ˆ ˆˆ
r ( cos 60) sin 60
22
a
aa ia j aij
r
= + + =+
4
ˆˆ
r0 aij =+
r
by substituting above value in the following formula
11 2 2 3 3 44
1 2 34
3
ˆ
r 0.95
4
mr m r m r mr
ai aj
m m mm
+ ++
= =+
+ ++
r r rr
r
So the location of centre of mass 
3
0.95 ,
4
aa
éù
êú
ëû
8. (d)
9. (d) m
1
 = 2kg, m
2
 = 4kg, 
12
2/ , 10/ v msv ms = =-
rr
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 20 4 10
0/
24
ms
´ -´
==
+
10. (a) As initially both the particles were at rest therefore
velocity of centre of mass was zero and there is no
external force on the system so speed of centre of mass
remains constant i.e. it should be equal to zero.
42
DPP/ P 14
11. (a) For translatory motion the force should be applied on
the centre of mass of the body, so we have to calculate the
location of centre of mass of 'T' shaped object.
Let mass of rod AB is m so the mass of rod CD will be 2m.
Let y
1
 is the centre of mass of rod AB and y
2
 is the centre of
mass of rod CD. We can consider that whole mass of the
rod is placed at their respective centre of mass i.e., mass m
is placed at y
1
 and mass 2 m is placed at y
2
.
 
B
A
x
C
y
2
y
1
l
l
y
D
Taking point 'C' at the origin, position vector of point y
1
and y
2
 can be written as 
1
r
ur
 = 
2
ˆˆ
2j,rj =
ur
l l , and m
1
 = m and
m
2
 = 2m
Position vector of centre of mass of the system
11 22
cm
12
ˆˆ
mr mr m2 j 2mj
r
m m m 2m
+ +
==
++
rr
r l l 
       = 
ˆ
4mj4
ˆ
j
3m3
=
l 
l
Hence the distance of centre of mass from 
4
C
3
= l
12. (a) Initial acceleration is zero of the system. So it will
always remain zero because there is no external force on
the system.
13. (b) According to figure let A is the origin and co-ordinates
of centre of mass be (x, y) then,
(x,y)
r
D C
x
B
2 kg
A
8 kg
4 kg 2 kg
y
11 223344
1 234
mx mx m x mx
x
m m mm
+ ++
=
+ ++
80 80
02 40
30
22
16 2
+´ +´ +
==
Similarly 
30
y
2
= so, 
22
r x y 30 cm = +=
14. (b) Linear density of the rod varies with distance
()
dm
Given dm dx
dx
= l \ =l
dx
x
Position of centre of mass
cm
dmx
x
dm
´
=
ò
ò
= 
3
0
3
0
() dxx
dx
l´
l
ò
ò
3
3 3
2
00
33
3
0
0
(2)
3
(2)
2
2
x
x x xdx
x
x dx
x
éù
+ +´ êú
êú
ëû
==
éù
+
+
êú
êú
ëû
ò
ò
= 
9 9 36 12
.
9
217
6
2
m
+
==
+
15. (c) Centre of mass lies always on the line that joins the
two particles.
For the combination cd and ab this line does not pass
through the origin.
For combination bd, initially it pass through the origin
but later on it moves toward negative x-axis.
 But for combination ac it will always pass through
origin. So we can say that centre of mass of this
combination will remain at origin.
Page 3


1. (c) m
1
 = 1, m
2
 = 35.5
1
, 
12
ˆ
0, 1.27 r ri
rr
==
11 22
12
mr mr
r
mm
rr
r ´
=
+
 Þ
35.5 1.27
ˆ
1 35.5
ri
´
=
+
r
35.5
ˆˆ
1.27 1.24
36.5
r ii = ´=
r
1
m 2
m
1.27Å
y
H
Cl
x
2. (d)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 3 3 2 12
2.4 m/s
235
´+´
= ==
+
3. (c) m
1
 =  12, m
2
 = 16
12
ˆˆ ˆˆ
0 0, 1.10 r i jr ij
r
= + =+
11 22
1
12
mr mr
r
mm
+
=
+
rr
r
C O
m
2
m
1
x
y
1
16 1.1
ˆˆ
0.63
28
r ii
´
==
r
i.e. 0.63Å from carbon atom.
4. (a)
1 1 2 2 33
1 23
cm
mv m v mv
v
m mm
++
=
++
r rr
r
= 
ˆ ˆˆ
20 10 30 10 50 10
100
i jk ´ +´ +´
\ 
ˆ ˆˆ
2 35
cm
v i jk =++
5. (c) We can assume that three particles of equal mass m are
placed at the corners of triangle.
12
ˆˆ ˆˆ
r 0i 0j,r bi 0j = + =+
u r ur
and 
3
ˆˆ
r 0i hj =+
ur
\ 
11 2 2 33
cm
1 23
m r m r mr
r
m mm
++
=
++
u r u r ur
uuu r
 
x
y
(b,0)
(0,0)
(0,h)
= 
bh
ˆˆ
ij
33
+
i.e. coordinates of centre of mass is 
bh
,
33
æö
ç÷
èø
6. (b)
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
       
224 10
7.3m/s
24
´ +´
==
+
7. (b) Let m
1
 = m, m
2
 = 2m, m
3
 = 3m, m
4
 = 4m
a
2m
m
3m
4m
x
y
60°
a cos 60°
a sin 60°
1
ˆˆ
00 r ij
r
=+
2
3
ˆ ˆˆ
r cos60 sin 60
22
aa
a ia j ij =+=+
r
3
33
ˆ ˆ ˆˆ
r ( cos 60) sin 60
22
a
aa ia j aij
r
= + + =+
4
ˆˆ
r0 aij =+
r
by substituting above value in the following formula
11 2 2 3 3 44
1 2 34
3
ˆ
r 0.95
4
mr m r m r mr
ai aj
m m mm
+ ++
= =+
+ ++
r r rr
r
So the location of centre of mass 
3
0.95 ,
4
aa
éù
êú
ëû
8. (d)
9. (d) m
1
 = 2kg, m
2
 = 4kg, 
12
2/ , 10/ v msv ms = =-
rr
11 22
12
cm
mv mv
v
mm
+
=
+
rr
r
2 20 4 10
0/
24
ms
´ -´
==
+
10. (a) As initially both the particles were at rest therefore
velocity of centre of mass was zero and there is no
external force on the system so speed of centre of mass
remains constant i.e. it should be equal to zero.
42
DPP/ P 14
11. (a) For translatory motion the force should be applied on
the centre of mass of the body, so we have to calculate the
location of centre of mass of 'T' shaped object.
Let mass of rod AB is m so the mass of rod CD will be 2m.
Let y
1
 is the centre of mass of rod AB and y
2
 is the centre of
mass of rod CD. We can consider that whole mass of the
rod is placed at their respective centre of mass i.e., mass m
is placed at y
1
 and mass 2 m is placed at y
2
.
 
B
A
x
C
y
2
y
1
l
l
y
D
Taking point 'C' at the origin, position vector of point y
1
and y
2
 can be written as 
1
r
ur
 = 
2
ˆˆ
2j,rj =
ur
l l , and m
1
 = m and
m
2
 = 2m
Position vector of centre of mass of the system
11 22
cm
12
ˆˆ
mr mr m2 j 2mj
r
m m m 2m
+ +
==
++
rr
r l l 
       = 
ˆ
4mj4
ˆ
j
3m3
=
l 
l
Hence the distance of centre of mass from 
4
C
3
= l
12. (a) Initial acceleration is zero of the system. So it will
always remain zero because there is no external force on
the system.
13. (b) According to figure let A is the origin and co-ordinates
of centre of mass be (x, y) then,
(x,y)
r
D C
x
B
2 kg
A
8 kg
4 kg 2 kg
y
11 223344
1 234
mx mx m x mx
x
m m mm
+ ++
=
+ ++
80 80
02 40
30
22
16 2
+´ +´ +
==
Similarly 
30
y
2
= so, 
22
r x y 30 cm = +=
14. (b) Linear density of the rod varies with distance
()
dm
Given dm dx
dx
= l \ =l
dx
x
Position of centre of mass
cm
dmx
x
dm
´
=
ò
ò
= 
3
0
3
0
() dxx
dx
l´
l
ò
ò
3
3 3
2
00
33
3
0
0
(2)
3
(2)
2
2
x
x x xdx
x
x dx
x
éù
+ +´ êú
êú
ëû
==
éù
+
+
êú
êú
ëû
ò
ò
= 
9 9 36 12
.
9
217
6
2
m
+
==
+
15. (c) Centre of mass lies always on the line that joins the
two particles.
For the combination cd and ab this line does not pass
through the origin.
For combination bd, initially it pass through the origin
but later on it moves toward negative x-axis.
 But for combination ac it will always pass through
origin. So we can say that centre of mass of this
combination will remain at origin.
DPP/ P 14
43
16. (b)
1011
111 3
cm
PQ PR PQ PR
x
´ +´ +´ +
==
++
and y
cm
 = 0
17. (a)
mg
R
2
R
1
Due to net force in downward direction and towards
left centre of mass will follow the path as shown in
figure.
18. (a) Initially both the particles were at rest so v
cm
 = 0. As
external force on the system is zero therefore velocity
of centre of mass remains unaffected.
19. (a)
11 22
0 mr mr
rr
+=
Þ
2
3
ˆ
150
44
mm
jr +=
r
Þ
2
ˆ
5 rj =-
r
i.e. larger fragment is at y = – 5 cm.
20. (b) Centre of mass is closer to massive part of the body
therefore the bottom piece of bat have larger mass.
21. (b) Initial position of centre of mass
1 1 22
12
cm
mx mx
r
mm
+
=
+
...(i)
1
m
2
m
2
x
1
x
d
If the particle of mass m
1
 is pushed towards the centre
of mass of the system through distance d and to keep
the centre of mass at the original position let second
particle displaced through distance d' away from the
centre of mass.
Now 
1 1 22
12
( ) ( ')
cm
mx d m xd
r
mm
+++
=
+
...(ii)
Equating (i) and  (ii)
1 1 22
12
mx mx
mm
++
+
1 1 22
12
( ) ( ') mx d m xd
mm
+++
=
+
By solving d' 
1
2
m
d
m
=-
Negative sign shows that particle m
2
 should be
displaced towards the centre of mass of the system.
22. (a) We know m
1
r
1
 =  m
2
r
2
 Þ m × r = constant 
1
r
m
\µ
23. (a) Depends on the distribution of mass in the body .
24. (a)
12
11 22
21
1 rm
mrmrr
rmm
= Þ = \µ
25. (b)
1 1 22
12 12
0 33
2
+ ´ +´
= ==
++
cm
ma ma m m gg
a
mm mm
26. (b)
v
m
m
0 m/s
0
l
By COE, 
2
2
131
22
mg
k mv
k
æö
=
ç÷
èø
2
9
3
mgm
vg
kk
==
03
22
´+
= ==
+
cm
m mv v gm
v
mmk
27. (a) By COE in CM-frame, 
22
11
22
ref
v kx m=
2
2
11
3
222
mm
g kx
k
æö
=
ç÷
èø
2
22
9
2
m
g kx
k
=  ;    
3
2
mg
x
k
=
28. (b) Statement-1 is True, Statement-2 is True; Statement-2
is NOT a correct explanation for Statement-1.
29. (a) Initially the electron and proton were at rest so their
centre of mass will be at rest. When they move towards
each other under mutual attraction then velocity of
centre of mass remains unaffected because external
force on the system is zero.
30. (d) The centre of mass of a system of particles depends
only on the masses of particles and the position of the
particles relative to one another. The location of
reference frame will not affect the location of centre of
mass.