Work, Energy and Power-2 Practice Questions - DPP for NEET

 Page 1


(1) (a) Let m
1
 and m
2
 be the masses of bullet and the rifleman
and v
1
 and v
2
 their respective velocities after the first shot.
Initially the rifleman and bullet are at rest, therefore initial
momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
   = m
1
v
1
 + m
2
v
2
or  v
2 
= 
2
1 1
m
v m
 = – 
kg 100
) s / m 800 ( ) kg 10 10 (
3 -
´
= – 0.08 m/s
V elocity acquired after 10 shots
= 10 v
2
 = 10 × (–0.08) = – 0.8 m/s
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite
to that of bullet.
(2) (c) Let the mass of block and bullet be M and m respectively
If v is the velocity of bullet and V is the velocity of block
with bullet embedded in it,
Now according to conservation of momentum ,
mv =  (M + m) V
(10×10 
–3
)(300) = (290×10
–3
 + 10 ×10
–3
) V or  V = 10 m/s
The kinetic energy just after impact is 
2
1
(M + m) V
2
, which
is lost due to work done on it by the force of friction F.
Since force of friction F = m (M+m)g and the work done is
given by Fd, we have
2
1
 (M + m) V
2
 = m (M + m) gd
or      m = 
2
1
 
gd
V
2
= 
2
1
 × 
) 15 ( ) 10 (
10
2
 = 
3
1
(3) (a) Let the initial velocity of the bullet of mass
m = 20 g = 0.020 kg be u and v the velocity with which each
plate moves.
The initial momentum of system (bullet + plate) = mu
m
4
m
2
m
Final momentum of system = m
1
v + (m
2
 + m) v
(Since bullet remains in 2nd plate)
\  According to principle of conservation of momentum
i.e.  mu = m
1
v + (m
2
 + m) v ,
i.e.  0.02u = 4v
or     u = 
02 .
4
 = 200 v ..........(1)
Let v
1
 be the velocity of the bullet as it comes out of plate
m
1
. Applying conservation of linear momentum for the
collision of bullet with plate m
2
.
i.e. mv
1
 = (m
2
 + m) v
    0.02 v
1
 = (2.98 + 0.02) v
i.e.  v
1
 = 
02 .
3
 v = 150 v .........(2)
Required percentage loss in initial velocity of bullet
u
v u
1
-
 × 100% = 
v 200
v 150 v 200 -
× 100 = 25%
(4) (a) Part (I) - The horizontal component of the momentum of
the bullet is equal to the momentum of the block with the
bullet
mu cos a = (M + m) V ..........(1)
Where V is the velocity of the block plus bullet embedded
in it.
Part (II) - As the block can move as a pendulum, the block
rises till its kinetic energy is converted into potential energy .
So, if the block rises upto a height h,
2
1
 (M + m) V
2
 = (M + m) gh ..........(2)
From (1) & (2)
h = 
2
m M
m
÷
ø
ö
ç
è
æ
+
. 
g 2
cos u
2 2
a
  = 
2
3
2
10 20
÷
÷
ø
ö
ç
ç
è
æ
´
-
.(200)
2 . 
cos
()()
2
30
2 10
o
 = 0.15 m
(5) (c)  Initial velocity of bullet, u
1
 = 500 m/s
Let v
1
 and v
2
 be the speeds of bullet and block after collision
u
1
m
1
m
2
v
2
0.1m
v
1
respectively then, 
2
1
 m
2
2
v = mgh
Þ v
2
 = gh 2 = 1 . 0 8 . 9 2 ´ ´ = 1.4 m/s
According to principle of conservation of linear momentum,
We have
   m
1
u
1
 + 0 = m
1
v
1
 + m
2 
v
2
or  0.01 × 500 = 0.01 v
1
 + 2 × 1.4 Þ v
1 
=
 
220 m/s
Page 2


(1) (a) Let m
1
 and m
2
 be the masses of bullet and the rifleman
and v
1
 and v
2
 their respective velocities after the first shot.
Initially the rifleman and bullet are at rest, therefore initial
momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
   = m
1
v
1
 + m
2
v
2
or  v
2 
= 
2
1 1
m
v m
 = – 
kg 100
) s / m 800 ( ) kg 10 10 (
3 -
´
= – 0.08 m/s
V elocity acquired after 10 shots
= 10 v
2
 = 10 × (–0.08) = – 0.8 m/s
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite
to that of bullet.
(2) (c) Let the mass of block and bullet be M and m respectively
If v is the velocity of bullet and V is the velocity of block
with bullet embedded in it,
Now according to conservation of momentum ,
mv =  (M + m) V
(10×10 
–3
)(300) = (290×10
–3
 + 10 ×10
–3
) V or  V = 10 m/s
The kinetic energy just after impact is 
2
1
(M + m) V
2
, which
is lost due to work done on it by the force of friction F.
Since force of friction F = m (M+m)g and the work done is
given by Fd, we have
2
1
 (M + m) V
2
 = m (M + m) gd
or      m = 
2
1
 
gd
V
2
= 
2
1
 × 
) 15 ( ) 10 (
10
2
 = 
3
1
(3) (a) Let the initial velocity of the bullet of mass
m = 20 g = 0.020 kg be u and v the velocity with which each
plate moves.
The initial momentum of system (bullet + plate) = mu
m
4
m
2
m
Final momentum of system = m
1
v + (m
2
 + m) v
(Since bullet remains in 2nd plate)
\  According to principle of conservation of momentum
i.e.  mu = m
1
v + (m
2
 + m) v ,
i.e.  0.02u = 4v
or     u = 
02 .
4
 = 200 v ..........(1)
Let v
1
 be the velocity of the bullet as it comes out of plate
m
1
. Applying conservation of linear momentum for the
collision of bullet with plate m
2
.
i.e. mv
1
 = (m
2
 + m) v
    0.02 v
1
 = (2.98 + 0.02) v
i.e.  v
1
 = 
02 .
3
 v = 150 v .........(2)
Required percentage loss in initial velocity of bullet
u
v u
1
-
 × 100% = 
v 200
v 150 v 200 -
× 100 = 25%
(4) (a) Part (I) - The horizontal component of the momentum of
the bullet is equal to the momentum of the block with the
bullet
mu cos a = (M + m) V ..........(1)
Where V is the velocity of the block plus bullet embedded
in it.
Part (II) - As the block can move as a pendulum, the block
rises till its kinetic energy is converted into potential energy .
So, if the block rises upto a height h,
2
1
 (M + m) V
2
 = (M + m) gh ..........(2)
From (1) & (2)
h = 
2
m M
m
÷
ø
ö
ç
è
æ
+
. 
g 2
cos u
2 2
a
  = 
2
3
2
10 20
÷
÷
ø
ö
ç
ç
è
æ
´
-
.(200)
2 . 
cos
()()
2
30
2 10
o
 = 0.15 m
(5) (c)  Initial velocity of bullet, u
1
 = 500 m/s
Let v
1
 and v
2
 be the speeds of bullet and block after collision
u
1
m
1
m
2
v
2
0.1m
v
1
respectively then, 
2
1
 m
2
2
v = mgh
Þ v
2
 = gh 2 = 1 . 0 8 . 9 2 ´ ´ = 1.4 m/s
According to principle of conservation of linear momentum,
We have
   m
1
u
1
 + 0 = m
1
v
1
 + m
2 
v
2
or  0.01 × 500 = 0.01 v
1
 + 2 × 1.4 Þ v
1 
=
 
220 m/s
38
DPP/ P 13
(6) (a) The rate of change of momentum is equal to force
F = 
dp
dt
= v 
dm
dt
 (Here v is constant)
Here v = 4 × 10
3 
m/s &
dm
dt
 = 50 × 10
–3 
kg/s
\ F = 4 × 10
3
 × 50 × 10 
–3
 = 200 N
(7) (a) Given that, Initial velocity = u
Final velocity = 
u
4
So by conservation of momentum, we have
1 × u + 0 = 1 × 
u
4
+ m × v
2 
Þ mv
2 
= 
3u
4
....... (1)
and by conservation of energy, we have
1
2
 × 1 × u
2
 + 0 =
1
2
 × 1 
2
u
4
æö
ç÷
èø
+ 
1
2
 m
2
2
v
or  
2
2
v
 
= 
15
16
u
2
.......
 
(2)
From equation (1) and (3),
2
2
2
2
(mv)
mv
= 
2
2
(9 /16)u
(15 /16)u
 or m = 0.6 kg
(8) (a) Initial momentum of the balls
= m × 9 + m × 0 = 9 m ........(1)
where m is the mass of each ball.
Let after collision their velocities are v
1
 and v
2 
respectively .
Final momentum of the balls after collision along the same
line = mv
1 
cos 30° + mv
2
 cos 30°
     = 
1
mv3
2
 + 
2
mv3
2
........(2)
According to law of conservation of momentum
9 m = 
1
mv3
2
 + 
2
mv3
2
   
3
2 9´
  = v
1
 + v
2
  .......(3)
Stationary ball
(a) Before collision
The initial momentum of the balls along perpendicular
direction = 0 .
Final momentum of balls along the perpendicular direction
= mv
1
 sin 30° – mv
2
 sin 30° = 
m
2
(v
1
 – v
2
)
Again by the law of conservation of momentum
(m/2) (v
1 
– v
2
) = 0
\ (v
1 
– v
2
) = 0 ......(4)
Solving equations (3) and (4), we have
v
1
  = 3 3 m/s and v
2
  = 3 3 m/s
30°
30°
v
2
v
1
According to law of conservation of energy
Energy before collision = Energy after collision
2
1
m
2
1
u + 
2
1
m
2
2
u = 
2
1
m
2
1
v + 
2
1
m
2
2
v
2
1
m (9)
2
 + 0 = 
2
1
m (3 3 )
2
  + 
2
1
m (3 3 )
2
  
81m
2
 = 
54m
2
L.H.S. # R.H.S.
i.e., energy is not conserved in this collision or this is a
case of inelastic collision.
(9) (a) The situation is shown in fig.
Let v
1
 and v
2
 be the velocities of two pieces after explosion.
Applying the law of conservation of energy, we have
30°
q
v
2
v
1
u = 50m/s
m = 8kg
m = 4kg
1
2
1
 (8) (50)
2
 + 15000 = 
2
1
 (D) 
2
1
v +
2
1
 (D) 
2
2
v
or 25000 = 2 (
2
1
v + 
2
2
v )     ......(1)
Applying the law of conservation of momentum along
x-axis and y-axis respectively , we get
8 (50) = 4 v
1
 cos q + v
2
 cos 30° ......(2)
and  0 = 4 v
1
 sin q
   = 4 v
2
 sin 30°
 
= 2 v
2
......(3)
or sin q =  
2
1
v
2v
......(4)
From eq. (2)
100 = v
1
 cos q  + v
2
 cos 30°
(10) (a) Let m be the mass of the rocket and v
r
 the relative
velocity of the gas ejecting from the rocket. Suppose the
fuel is burnt at a rate (dm/dt) to provide the rocket an
acceleration a.
Page 3


(1) (a) Let m
1
 and m
2
 be the masses of bullet and the rifleman
and v
1
 and v
2
 their respective velocities after the first shot.
Initially the rifleman and bullet are at rest, therefore initial
momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
   = m
1
v
1
 + m
2
v
2
or  v
2 
= 
2
1 1
m
v m
 = – 
kg 100
) s / m 800 ( ) kg 10 10 (
3 -
´
= – 0.08 m/s
V elocity acquired after 10 shots
= 10 v
2
 = 10 × (–0.08) = – 0.8 m/s
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite
to that of bullet.
(2) (c) Let the mass of block and bullet be M and m respectively
If v is the velocity of bullet and V is the velocity of block
with bullet embedded in it,
Now according to conservation of momentum ,
mv =  (M + m) V
(10×10 
–3
)(300) = (290×10
–3
 + 10 ×10
–3
) V or  V = 10 m/s
The kinetic energy just after impact is 
2
1
(M + m) V
2
, which
is lost due to work done on it by the force of friction F.
Since force of friction F = m (M+m)g and the work done is
given by Fd, we have
2
1
 (M + m) V
2
 = m (M + m) gd
or      m = 
2
1
 
gd
V
2
= 
2
1
 × 
) 15 ( ) 10 (
10
2
 = 
3
1
(3) (a) Let the initial velocity of the bullet of mass
m = 20 g = 0.020 kg be u and v the velocity with which each
plate moves.
The initial momentum of system (bullet + plate) = mu
m
4
m
2
m
Final momentum of system = m
1
v + (m
2
 + m) v
(Since bullet remains in 2nd plate)
\  According to principle of conservation of momentum
i.e.  mu = m
1
v + (m
2
 + m) v ,
i.e.  0.02u = 4v
or     u = 
02 .
4
 = 200 v ..........(1)
Let v
1
 be the velocity of the bullet as it comes out of plate
m
1
. Applying conservation of linear momentum for the
collision of bullet with plate m
2
.
i.e. mv
1
 = (m
2
 + m) v
    0.02 v
1
 = (2.98 + 0.02) v
i.e.  v
1
 = 
02 .
3
 v = 150 v .........(2)
Required percentage loss in initial velocity of bullet
u
v u
1
-
 × 100% = 
v 200
v 150 v 200 -
× 100 = 25%
(4) (a) Part (I) - The horizontal component of the momentum of
the bullet is equal to the momentum of the block with the
bullet
mu cos a = (M + m) V ..........(1)
Where V is the velocity of the block plus bullet embedded
in it.
Part (II) - As the block can move as a pendulum, the block
rises till its kinetic energy is converted into potential energy .
So, if the block rises upto a height h,
2
1
 (M + m) V
2
 = (M + m) gh ..........(2)
From (1) & (2)
h = 
2
m M
m
÷
ø
ö
ç
è
æ
+
. 
g 2
cos u
2 2
a
  = 
2
3
2
10 20
÷
÷
ø
ö
ç
ç
è
æ
´
-
.(200)
2 . 
cos
()()
2
30
2 10
o
 = 0.15 m
(5) (c)  Initial velocity of bullet, u
1
 = 500 m/s
Let v
1
 and v
2
 be the speeds of bullet and block after collision
u
1
m
1
m
2
v
2
0.1m
v
1
respectively then, 
2
1
 m
2
2
v = mgh
Þ v
2
 = gh 2 = 1 . 0 8 . 9 2 ´ ´ = 1.4 m/s
According to principle of conservation of linear momentum,
We have
   m
1
u
1
 + 0 = m
1
v
1
 + m
2 
v
2
or  0.01 × 500 = 0.01 v
1
 + 2 × 1.4 Þ v
1 
=
 
220 m/s
38
DPP/ P 13
(6) (a) The rate of change of momentum is equal to force
F = 
dp
dt
= v 
dm
dt
 (Here v is constant)
Here v = 4 × 10
3 
m/s &
dm
dt
 = 50 × 10
–3 
kg/s
\ F = 4 × 10
3
 × 50 × 10 
–3
 = 200 N
(7) (a) Given that, Initial velocity = u
Final velocity = 
u
4
So by conservation of momentum, we have
1 × u + 0 = 1 × 
u
4
+ m × v
2 
Þ mv
2 
= 
3u
4
....... (1)
and by conservation of energy, we have
1
2
 × 1 × u
2
 + 0 =
1
2
 × 1 
2
u
4
æö
ç÷
èø
+ 
1
2
 m
2
2
v
or  
2
2
v
 
= 
15
16
u
2
.......
 
(2)
From equation (1) and (3),
2
2
2
2
(mv)
mv
= 
2
2
(9 /16)u
(15 /16)u
 or m = 0.6 kg
(8) (a) Initial momentum of the balls
= m × 9 + m × 0 = 9 m ........(1)
where m is the mass of each ball.
Let after collision their velocities are v
1
 and v
2 
respectively .
Final momentum of the balls after collision along the same
line = mv
1 
cos 30° + mv
2
 cos 30°
     = 
1
mv3
2
 + 
2
mv3
2
........(2)
According to law of conservation of momentum
9 m = 
1
mv3
2
 + 
2
mv3
2
   
3
2 9´
  = v
1
 + v
2
  .......(3)
Stationary ball
(a) Before collision
The initial momentum of the balls along perpendicular
direction = 0 .
Final momentum of balls along the perpendicular direction
= mv
1
 sin 30° – mv
2
 sin 30° = 
m
2
(v
1
 – v
2
)
Again by the law of conservation of momentum
(m/2) (v
1 
– v
2
) = 0
\ (v
1 
– v
2
) = 0 ......(4)
Solving equations (3) and (4), we have
v
1
  = 3 3 m/s and v
2
  = 3 3 m/s
30°
30°
v
2
v
1
According to law of conservation of energy
Energy before collision = Energy after collision
2
1
m
2
1
u + 
2
1
m
2
2
u = 
2
1
m
2
1
v + 
2
1
m
2
2
v
2
1
m (9)
2
 + 0 = 
2
1
m (3 3 )
2
  + 
2
1
m (3 3 )
2
  
81m
2
 = 
54m
2
L.H.S. # R.H.S.
i.e., energy is not conserved in this collision or this is a
case of inelastic collision.
(9) (a) The situation is shown in fig.
Let v
1
 and v
2
 be the velocities of two pieces after explosion.
Applying the law of conservation of energy, we have
30°
q
v
2
v
1
u = 50m/s
m = 8kg
m = 4kg
1
2
1
 (8) (50)
2
 + 15000 = 
2
1
 (D) 
2
1
v +
2
1
 (D) 
2
2
v
or 25000 = 2 (
2
1
v + 
2
2
v )     ......(1)
Applying the law of conservation of momentum along
x-axis and y-axis respectively , we get
8 (50) = 4 v
1
 cos q + v
2
 cos 30° ......(2)
and  0 = 4 v
1
 sin q
   = 4 v
2
 sin 30°
 
= 2 v
2
......(3)
or sin q =  
2
1
v
2v
......(4)
From eq. (2)
100 = v
1
 cos q  + v
2
 cos 30°
(10) (a) Let m be the mass of the rocket and v
r
 the relative
velocity of the gas ejecting from the rocket. Suppose the
fuel is burnt at a rate (dm/dt) to provide the rocket an
acceleration a.
DPP/ P 13
39
Then  a = 
r
v
m
 
dm
dt
æö
ç÷
èø
 – g     ......(1)
Here v
r
 = 250 m/s, m = 500 kg, g = 10 m/s
2
 and a = 20 m/s
2
Now from (1)   
dm
dt
 = 
r
m
v
(a + g)
                 = 
500
250
  (20 + 10)  = 60 kg/s
(11) (a) Let m
1
 and m
2
 be the masses of electron and hydrogen
atom respectively. If u
1
 and v
1
 be the initial and final
velocities of electron, then initial kinetic energy of electron
K
i
 = 
1
2
æö
ç÷
èø
 m
2
1
u
Final kinetic energy of electron  K
f
  = 
1
2
æö
ç÷
èø
 m
2
1
v
Fractional decrease in K.E.,
if
i
KK
K
-
 = 
2
1
2
1
v
1
u
-
.........(1)
For such a collision, we have
v
1
 = 
12
12
mm
mm
æö -
ç÷
+ èø
u
1
\    
1
1
v
u
 = 
12
12
mm
mm
æö -
ç÷
+ èø
.........(2)
From eqs. (1) and (2) we have
 
if
i
KK
K
-
 = 1 – 
2
12
12
mm
mm
æö -
ç÷
+ èø
  = 
12
2
12
4mm
(m m) +
or  
if
i
KK
K
-
  = 
21
2
1
2
4(m /m)
(1 m /m) +
 = 
2
4 1850
(1 1850)
´
+
     = 0.00217 = 0.217%
(12) (c)
m
m
2m
v/ 2
Now the total energy released in the explosion
= 
1
2
mv
2 
+ 
1
2
mv
2 
+ 
1
2
2m
2
v
2
æö
ç÷
èø
 =  
3
2
 mv
2
(13) (a) Let the speed of the body before explosion be u. After
explosion, if the two parts move with velocities u
1
 and u
2
in the same direction, then according to conservation of
momentum,
a Mu
1
 + (1 – a) M u
2
 = Mu
The kinetic energy T liberated during explosion  is given
by T = 
1
2
a M
2
1
u + 
1
2
(1 – a) M
2
2
u – 
1
2
 Mu
2
  = 
1
2
a M
2
1
u + 
1
2
 (1 – a) M
2
2
u – 
1
2M
[ a Mu
1
 + (1 – a) Mu
2
]
2
  = 
1
2
M a (1 – a) [
2
1
u + 
2
2
u
 
– 2 u
1
 u
2
]
    (u
1 
– u
2
)
2 
= 
2T
(1 )M a -a
Þ  (u
1 
– u
2
) =
2T
(1 )M a -a
(14) (a) The situation is shown in fig.
Let A and B be two pieces of equal mass (1/5 kg)  which fly
off perpendicular to a each other with equal velocity
(30 m/sec)
Momentum of A or B = (1/5 × 30)
B
1
kg
5
m =
v = 30 m/s
1
kg
5
m =
v = 30 m/s
3
kg
5
m =
C
A
\ Resultant momentum
   =
22
{(1/ 5) 30} {(1/ 5) 30} ´ +´  = 6 2 kg m/sec
along the bisector of Ð AOB
(3/5) × v = 6 2 Þ v = 10 2 m/sec
(15) (c) The situation is shown in fig.
Equating the total initial and final momentum along each
axis, we get
mv
1
 + 0 = (M + m) v' cos q
 .........
(A)
0 + Mv
2
 = (M + m) v' sin q
 .........
(B)
Squaring and adding eq. (A) and (B), we get
(mv
1
)
2
 + (Mv
2
)
2
 = (M + m)
2
 v'
2 .........
(C)
Page 4


(1) (a) Let m
1
 and m
2
 be the masses of bullet and the rifleman
and v
1
 and v
2
 their respective velocities after the first shot.
Initially the rifleman and bullet are at rest, therefore initial
momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
   = m
1
v
1
 + m
2
v
2
or  v
2 
= 
2
1 1
m
v m
 = – 
kg 100
) s / m 800 ( ) kg 10 10 (
3 -
´
= – 0.08 m/s
V elocity acquired after 10 shots
= 10 v
2
 = 10 × (–0.08) = – 0.8 m/s
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite
to that of bullet.
(2) (c) Let the mass of block and bullet be M and m respectively
If v is the velocity of bullet and V is the velocity of block
with bullet embedded in it,
Now according to conservation of momentum ,
mv =  (M + m) V
(10×10 
–3
)(300) = (290×10
–3
 + 10 ×10
–3
) V or  V = 10 m/s
The kinetic energy just after impact is 
2
1
(M + m) V
2
, which
is lost due to work done on it by the force of friction F.
Since force of friction F = m (M+m)g and the work done is
given by Fd, we have
2
1
 (M + m) V
2
 = m (M + m) gd
or      m = 
2
1
 
gd
V
2
= 
2
1
 × 
) 15 ( ) 10 (
10
2
 = 
3
1
(3) (a) Let the initial velocity of the bullet of mass
m = 20 g = 0.020 kg be u and v the velocity with which each
plate moves.
The initial momentum of system (bullet + plate) = mu
m
4
m
2
m
Final momentum of system = m
1
v + (m
2
 + m) v
(Since bullet remains in 2nd plate)
\  According to principle of conservation of momentum
i.e.  mu = m
1
v + (m
2
 + m) v ,
i.e.  0.02u = 4v
or     u = 
02 .
4
 = 200 v ..........(1)
Let v
1
 be the velocity of the bullet as it comes out of plate
m
1
. Applying conservation of linear momentum for the
collision of bullet with plate m
2
.
i.e. mv
1
 = (m
2
 + m) v
    0.02 v
1
 = (2.98 + 0.02) v
i.e.  v
1
 = 
02 .
3
 v = 150 v .........(2)
Required percentage loss in initial velocity of bullet
u
v u
1
-
 × 100% = 
v 200
v 150 v 200 -
× 100 = 25%
(4) (a) Part (I) - The horizontal component of the momentum of
the bullet is equal to the momentum of the block with the
bullet
mu cos a = (M + m) V ..........(1)
Where V is the velocity of the block plus bullet embedded
in it.
Part (II) - As the block can move as a pendulum, the block
rises till its kinetic energy is converted into potential energy .
So, if the block rises upto a height h,
2
1
 (M + m) V
2
 = (M + m) gh ..........(2)
From (1) & (2)
h = 
2
m M
m
÷
ø
ö
ç
è
æ
+
. 
g 2
cos u
2 2
a
  = 
2
3
2
10 20
÷
÷
ø
ö
ç
ç
è
æ
´
-
.(200)
2 . 
cos
()()
2
30
2 10
o
 = 0.15 m
(5) (c)  Initial velocity of bullet, u
1
 = 500 m/s
Let v
1
 and v
2
 be the speeds of bullet and block after collision
u
1
m
1
m
2
v
2
0.1m
v
1
respectively then, 
2
1
 m
2
2
v = mgh
Þ v
2
 = gh 2 = 1 . 0 8 . 9 2 ´ ´ = 1.4 m/s
According to principle of conservation of linear momentum,
We have
   m
1
u
1
 + 0 = m
1
v
1
 + m
2 
v
2
or  0.01 × 500 = 0.01 v
1
 + 2 × 1.4 Þ v
1 
=
 
220 m/s
38
DPP/ P 13
(6) (a) The rate of change of momentum is equal to force
F = 
dp
dt
= v 
dm
dt
 (Here v is constant)
Here v = 4 × 10
3 
m/s &
dm
dt
 = 50 × 10
–3 
kg/s
\ F = 4 × 10
3
 × 50 × 10 
–3
 = 200 N
(7) (a) Given that, Initial velocity = u
Final velocity = 
u
4
So by conservation of momentum, we have
1 × u + 0 = 1 × 
u
4
+ m × v
2 
Þ mv
2 
= 
3u
4
....... (1)
and by conservation of energy, we have
1
2
 × 1 × u
2
 + 0 =
1
2
 × 1 
2
u
4
æö
ç÷
èø
+ 
1
2
 m
2
2
v
or  
2
2
v
 
= 
15
16
u
2
.......
 
(2)
From equation (1) and (3),
2
2
2
2
(mv)
mv
= 
2
2
(9 /16)u
(15 /16)u
 or m = 0.6 kg
(8) (a) Initial momentum of the balls
= m × 9 + m × 0 = 9 m ........(1)
where m is the mass of each ball.
Let after collision their velocities are v
1
 and v
2 
respectively .
Final momentum of the balls after collision along the same
line = mv
1 
cos 30° + mv
2
 cos 30°
     = 
1
mv3
2
 + 
2
mv3
2
........(2)
According to law of conservation of momentum
9 m = 
1
mv3
2
 + 
2
mv3
2
   
3
2 9´
  = v
1
 + v
2
  .......(3)
Stationary ball
(a) Before collision
The initial momentum of the balls along perpendicular
direction = 0 .
Final momentum of balls along the perpendicular direction
= mv
1
 sin 30° – mv
2
 sin 30° = 
m
2
(v
1
 – v
2
)
Again by the law of conservation of momentum
(m/2) (v
1 
– v
2
) = 0
\ (v
1 
– v
2
) = 0 ......(4)
Solving equations (3) and (4), we have
v
1
  = 3 3 m/s and v
2
  = 3 3 m/s
30°
30°
v
2
v
1
According to law of conservation of energy
Energy before collision = Energy after collision
2
1
m
2
1
u + 
2
1
m
2
2
u = 
2
1
m
2
1
v + 
2
1
m
2
2
v
2
1
m (9)
2
 + 0 = 
2
1
m (3 3 )
2
  + 
2
1
m (3 3 )
2
  
81m
2
 = 
54m
2
L.H.S. # R.H.S.
i.e., energy is not conserved in this collision or this is a
case of inelastic collision.
(9) (a) The situation is shown in fig.
Let v
1
 and v
2
 be the velocities of two pieces after explosion.
Applying the law of conservation of energy, we have
30°
q
v
2
v
1
u = 50m/s
m = 8kg
m = 4kg
1
2
1
 (8) (50)
2
 + 15000 = 
2
1
 (D) 
2
1
v +
2
1
 (D) 
2
2
v
or 25000 = 2 (
2
1
v + 
2
2
v )     ......(1)
Applying the law of conservation of momentum along
x-axis and y-axis respectively , we get
8 (50) = 4 v
1
 cos q + v
2
 cos 30° ......(2)
and  0 = 4 v
1
 sin q
   = 4 v
2
 sin 30°
 
= 2 v
2
......(3)
or sin q =  
2
1
v
2v
......(4)
From eq. (2)
100 = v
1
 cos q  + v
2
 cos 30°
(10) (a) Let m be the mass of the rocket and v
r
 the relative
velocity of the gas ejecting from the rocket. Suppose the
fuel is burnt at a rate (dm/dt) to provide the rocket an
acceleration a.
DPP/ P 13
39
Then  a = 
r
v
m
 
dm
dt
æö
ç÷
èø
 – g     ......(1)
Here v
r
 = 250 m/s, m = 500 kg, g = 10 m/s
2
 and a = 20 m/s
2
Now from (1)   
dm
dt
 = 
r
m
v
(a + g)
                 = 
500
250
  (20 + 10)  = 60 kg/s
(11) (a) Let m
1
 and m
2
 be the masses of electron and hydrogen
atom respectively. If u
1
 and v
1
 be the initial and final
velocities of electron, then initial kinetic energy of electron
K
i
 = 
1
2
æö
ç÷
èø
 m
2
1
u
Final kinetic energy of electron  K
f
  = 
1
2
æö
ç÷
èø
 m
2
1
v
Fractional decrease in K.E.,
if
i
KK
K
-
 = 
2
1
2
1
v
1
u
-
.........(1)
For such a collision, we have
v
1
 = 
12
12
mm
mm
æö -
ç÷
+ èø
u
1
\    
1
1
v
u
 = 
12
12
mm
mm
æö -
ç÷
+ èø
.........(2)
From eqs. (1) and (2) we have
 
if
i
KK
K
-
 = 1 – 
2
12
12
mm
mm
æö -
ç÷
+ èø
  = 
12
2
12
4mm
(m m) +
or  
if
i
KK
K
-
  = 
21
2
1
2
4(m /m)
(1 m /m) +
 = 
2
4 1850
(1 1850)
´
+
     = 0.00217 = 0.217%
(12) (c)
m
m
2m
v/ 2
Now the total energy released in the explosion
= 
1
2
mv
2 
+ 
1
2
mv
2 
+ 
1
2
2m
2
v
2
æö
ç÷
èø
 =  
3
2
 mv
2
(13) (a) Let the speed of the body before explosion be u. After
explosion, if the two parts move with velocities u
1
 and u
2
in the same direction, then according to conservation of
momentum,
a Mu
1
 + (1 – a) M u
2
 = Mu
The kinetic energy T liberated during explosion  is given
by T = 
1
2
a M
2
1
u + 
1
2
(1 – a) M
2
2
u – 
1
2
 Mu
2
  = 
1
2
a M
2
1
u + 
1
2
 (1 – a) M
2
2
u – 
1
2M
[ a Mu
1
 + (1 – a) Mu
2
]
2
  = 
1
2
M a (1 – a) [
2
1
u + 
2
2
u
 
– 2 u
1
 u
2
]
    (u
1 
– u
2
)
2 
= 
2T
(1 )M a -a
Þ  (u
1 
– u
2
) =
2T
(1 )M a -a
(14) (a) The situation is shown in fig.
Let A and B be two pieces of equal mass (1/5 kg)  which fly
off perpendicular to a each other with equal velocity
(30 m/sec)
Momentum of A or B = (1/5 × 30)
B
1
kg
5
m =
v = 30 m/s
1
kg
5
m =
v = 30 m/s
3
kg
5
m =
C
A
\ Resultant momentum
   =
22
{(1/ 5) 30} {(1/ 5) 30} ´ +´  = 6 2 kg m/sec
along the bisector of Ð AOB
(3/5) × v = 6 2 Þ v = 10 2 m/sec
(15) (c) The situation is shown in fig.
Equating the total initial and final momentum along each
axis, we get
mv
1
 + 0 = (M + m) v' cos q
 .........
(A)
0 + Mv
2
 = (M + m) v' sin q
 .........
(B)
Squaring and adding eq. (A) and (B), we get
(mv
1
)
2
 + (Mv
2
)
2
 = (M + m)
2
 v'
2 .........
(C)
40
DPP/ P 13
(M+m)v¢
q
mv
1
Mv
2
x
y
A
The final momentum
P = (M + m) v'
 
=
22
12
[(mv ) (Mv ) ] +
[form eqn. (3)]
Dividing eqn. (2) by eqn. (1), we have
tan q = 
2
1
Mv
mv
   or q = tan
–1
2
1
Mv
mv
æö
ç÷
èø
(16) (a) Let the angle of reflection be q' and the magnitude of
velocity after collision be v'. As there is no force parallel to
the wall, the component of velocity parallel to the surface
remains unchanged.
Therefore, v' sin q ' = v sin q ......(1)
As the coefficient of restitution is e, for perpendicular
component of velocity
V elocity of separation = e x velocity of approach
–(v'
 
cos q ' – 0) = –e (v cos q – 0) ......(2)
From (1) and (2)
     v' = 
2 22
v sin e cos q+q
and     tan q ' =  tan q/e
(17) (a) The fraction of energy lost is given by ,
E
E
D
 = 
mg(h h ')
mgh
-
 = 
h h'
h
-
given that, h = 2 meter and h' = 1.5 meter
\  
E
E
D
 = 
2 1.51
24
-
=
(18) (a) A bullet is fired from the gun. The gun recoils, the kinetic
energy of the recoil shall be less than the kinetic energy of
the bullet.
(19) (a) Conservation of linear momentum is equivalent to
Newton's second law of motion
(20) (a) In an inelastic  collision momentum is conserved but
kinetic energy is not.
(21) (a) Inelastic collision is the collision of electron and
positron to an inhilate each other.
(22) (a) Total kinetic energy is not conserved in inelastic
collisions but momentum is conserved
(23) (a) (1) when m
1
 = m
2
 and m
2
 is stationary , there is maximum
transfer of kinetic energy in head an collision
(2) when m
1
 = m
2
 and m
2
 is stationary , there is maximum
transfer of momentum in head on collision
(3) when m
1
 >> m
2
 and m
2
 is stationary, after head on
collision m
2
 moves with twice the velocity of m
1
.
(24) (a) Momentum remains conserved
(25) (a) Speed of particle after the collision
2
15
3 25
43
æö
= ´+
ç÷
èø
 = 5.036 m /s
(26) (b) Speed of the sphere just after collision 
30
43
= m/s
(27) (a) Angular speed of sphere is zero as impulse due to
collision passes through centre of sphere.
(28) (c) When e = 0, velocity of separation along common normal
zero, but there may be relative velocity along common
tangent.
(29) (c) Statement – 1 is false but statement – 2 is true.
(30) (d) Momentum remains constant before, during and after
the collision but KE does not remain constant during
the collision as the energy gets converted into elastic
potential energy due to deformation.
t.me/Ebooks_Encyclopedia27.    t.me/Magazines4all