Chapter 11 Work, Energy and Power Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 11 Work, Energy and Power Part 3 – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-11-Work-Energy-and-Power-Part-3.pdf

 

 

 

Problem 76.      A particle of mass m moving with horizontal speed 6 m/sec. If m<<M then for one dimensional elastic collision, the speed of lighter particle after collision will be                                                          [MP PMT 2003]

(a) 2 m/sec in original direction                                   (b) 2 m/sec opposite to the original direction

(c) 4 m/sec opposite to the original direction                  (d) 4 m/sec in original direction

æ m1 – m2 ö             2m2u2

 

Solution : (a)

v1 = ç m   m   ÷ u1 + m   m

 

è   1          2 ø              1         2

Substituting m1 = 0, v1 = –u1 + 2u2

 

Þ v1 = – 6 + 2(4)

= 2m/s

 

i.e. the lighter particle will move in original direction with the speed of 2 m/s.

Problem 77.      A body of mass m moving with velocity v makes a head-on collision with another body of mass 2m which is initially at rest. The loss of kinetic energy of the colliding body (mass m) is                          [MP PMT 1996; RPET 1999; AIIMS 2003]

 

 

(a)

1 of its initial kinetic energy                              (b)

2

1 of its initial kinetic energy

9

 

 

(c)

8 of its initial kinetic energy                              (d)

9

1 of its initial kinetic energy

4

 

DK             æ m – m   ö 2

æ m – 2m ö 2

æ 1 ö 2

 

Solution : (c)      Loss of kinetic energy of the colliding body

= 1 – ç    1        2 ÷

= 1 – ç

÷  = 1 – ç  ÷

 

K              è m1 + m2 ø

è m + 2m ø

è 3 ø

 

 

DK = æ1 – 1 ö K = 8 K      \ Loss of kinetic energy is 8 of its initial kinetic energy.

ç           ÷

9
9
9

è           ø

Problem 78. A ball of mass m moving with velocity V, makes a head on elastic collision with a ball of the same mass moving with velocity 2V towards it. Taking direction of V as positive velocities of the two balls after collision are                                                                                                                                        [MP PMT 2002]

  • V and 2V (b) 2V and – V                   (c) V and 2V                      (d) – 2V and V

Solution : (d) Initial velocities of balls are +V and – 2V respectively and we know that for given condition velocities get interchanged after collision. So the velocities of two balls after collision are – 2V and V respectively.

Problem 79.      Consider the following statements

Assertion (A) : In an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of collision of the balls (i.e., when they are in contact)

Reason (R) : Energy spent against friction does not follow the law of conservation of energy of these statements                                                                                                                            [AIIMS 2002]

  • Both A and R are true and the R is a correct explanation of A
  • Both A and R are true but the R is not a correct explanation of the A
  • A is true but the R is false
  • Both A and R are false

Solution : (d) (i) When they are in contact some part of kinetic energy may convert in potential energy so it is not conserved during the short time of collision. (ii) Law of conservation of energy is always true.

 

 

 

Problem 80.      A big ball of mass M, moving with velocity u strikes a small ball of mass m, which is at rest. Finally small ball attains velocity u and big ball v. Then what is the value of v                                                                                                [RPET 2001]

 

 

(a)

Mm u M + m

(b)

m      u M + m

(c)

2m u M + m

(d)

M      u M + m

 

æ m1 – m2 ö           æ M m ö

Solution : (a)      From the standard equation v1 = ç m m ÷ u1 = ç M m ÷ u .

è   1         2 ø          è              ø

 

Problem 81.      A car of mass

400kg

and travelling at 72 kmph crashes into a truck of mass

4000kg

and travelling at 9

 

kmph, in the same direction. The car bounces back at a speed of 18 kmph. The speed of the truck after the impact is                                                                                                                             [EAMCET (Engg.) 1997]

  • 9 kmph (b) 18 kmph                        (c) 27 kmph                           (d) 36 kmph Solution : (b)      By the law of conservation of linear momentum m1u1 + m2u2 = m1v1 + m2v2

Þ     400 ´ 72 + 4000 ´ 9 = 400 ´ (-18) + 4000 ´ v2           Þ v2 = 18 km / h .

Problem 82.      A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m

at rest. After collision their final velocities are V and v respectively. The value of v is                           [MP PET 1995]

 

 

(a)

2uM m

2um M

 

æ m2 – m1 ö

 

 

 

2m1u1

2u 1 + m

M

2u 1 + M

m

 

Solution : (c)      Final velocity of the target v2 = ç m m

÷u2 + m   + m

 

è   1          2 ø             1         2

As initially target is at rest so by substituting u2 = 0 we get v2 = 2Mu  =  2u  .

 

M + m

1 + m

M

 

Problem 83. A sphere of mass 0.1 kg is attached to a cord of 1m length. Starting from the height of its point of suspension this sphere hits a block of same mass at rest on a frictionless table, If the impact is elastic, then the kinetic energy of the block after the collision is                                                                                                                             [RPET 1991]

 

  • 1 J
  • 10 J
  • 1 J
  • 5 J

Solution : (a)      As two blocks are of same mass and the collision is perfectly elastic therefore their velocities gets interchanged

i.e. the block A comes into rest and complete kinetic energy transferred to block B.

Now kinetic energy of block B after collision = Kinetic energy of block A before collision

= Potential energy of block A at the original height

= mgh = 0.1 ´ 10 ´ 1 = 1 J.

Problem 84. A ball moving horizontally with speed v strikes the bob of a simple pendulum at rest. The mass of the bob is equal to that of the ball. If the collision is elastic the bob will rise to a height

 

(a) v 2

g

(b)

v 2                                  (c)

2g

v 2                                       (d)  v 2

4 g                                              8g

 

 

 

Solution : (b)      Total kinetic energy of the ball will transfer to the bob of simple pendulum. Let it rises to height ‘h’ by the law of conservation of energy.

1 mv 2 = mgh

2

\ h = v 2

2g

 

Problem 85.      A moving body with a mass m1 strikes a stationary body of mass m2. The masses m1 and m2

should be in the

 

ratio m1

m2

so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then

 

the ratio m1 is

m2

(a) 1/ 25                          (b) 1/5                          (c) 5                               (d) 25

 

æ m1 – m2 ö           2m2u2

 

æ m1 – m2 ö

 

æ          u1 ö

 

Solution : (c)

v1 = ç m + m ÷u1 + m + m

= ç m m ÷u1

[As u2 = 0 and çv1 = 1.5 ÷ given]

 

è   1        2 ø            1

u1       æ m1 – m2 ö

 

2      è   1         2 ø                                                  è                 ø

m1

 

Þ             = ç

1.5      m + m

÷u1 Þ m1 + m2 = 1.5(m1 – m2 ) Þ m

= 5 .

 

è   1         2 ø                                                                    2

Problem 86.      Six identical balls are lined in a straight groove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide with the row of 6 balls from left. What will happen

  • One ball from the right rolls out with a speed 2v and the remaining balls will remain at rest
  • Two balls from the right roll out with speed v each and the remaining balls will remain stationary
  • All the six balls in the row will roll out with speed v/6 each and the two colliding balls will come to rest
  • The colliding balls will come to rest and no ball rolls out from right

Solution : (b)      Only this condition satisfies the law of conservation of linear momentum.

Problem 87.      A moving mass of 8 kg collides elastically with a stationary mass of 2 kg. If E be the initial kinetic energy of the mass, the kinetic energy left with it after collision will be

(a) 0.80 E                                 (b) 0.64 E                            (c) 0.36 E                                (d) 0.08 E

DK      æ m   – m   ö 2                     æ 8 – 2 ö 2            9

 

Solution : (c)      Kinetic energy retained by projectile

= ç   1          2 ÷

Þ DK = ç

÷  E = =

E = 0.36E .

 

K       è m1 + m2 ø                        è 8 + 2 ø                  25

Problem 88.      A neutron travelling with a velocity v and K.E. E collides perfectly elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by neutron is

 

æ A – 1 ö 2

æ A + 1 ö 2

æ A – 1 ö 2

æ A + 1 ö 2

 

(a)

ç A + 1 ÷

(b)

ç A – 1 ÷

  • ç A   ÷
  • ç A   ÷

 

è            ø                                     è            ø                                 è            ø                                     è            ø

 

 

 

DK      æ m – m   ö 2

 

Solution : (a)      Fraction of kinetic energy retained by projectile

= ç   1          2 ÷

K       è m1 + m2 ø

DK

æ 1 – A ö 2

æ A – 1 ö 2

 

Mass of neutron (m1) = 1 and Mass of atom (m2) = A                    \

K   = ç 1 + A ÷

or ç A + 1 ÷ .

 

è            ø          è            ø

Problem 89.      A neutron with 0.6MeV kinetic energy directly collides with a stationary carbon nucleus (mass number 12).

The kinetic energy of carbon nucleus after the collision is

  • 7 MeV (b) 0.17 MeV                      (c) 17 MeV                              (d) Zero

DK       é      æ m – m   ö 2 ù

 

Solution : (b)      Kinetic energy transferred to stationary target (carbon nucleus)

= ê1 – ç  1        2 ÷ ú

 

 

DK       é      æ 1 – 12 ö 2 ù         é

 

121 ù         48

K        êë     è m1 + m2 ø úû

48

 

K = ê1 – ç 1 + 12 ÷

ú = ê1 – 169 ú = 169

\ DK =           ´ (0.6MeV) = 0.17MeV.

169

 

ëê     è             ø úû     ë               û

Problem 90. A body of mass m moving along a straight line collides with a body of mass nm which is also moving with a velocity kv in the same direction. If the first body comes to rest after the collision, then the velocity of second body after the collision would be

 

(a)

nv

 

(1 + nk)

(b)

nv

 

(1 – nk)

(c)

(1 – nk)v n

(d)

(1 + nk)v n

 

Solution : (d)     Initial momentum = mv + nm(kv)            and      final momentum = 0 + nm V

By the conservation of momentum, mv + nm(kv) = 0 + nm V

 

 

Þ v + nkv = nV

Þ nV = (1 + nk)v Þ V = (1 + nk)v

n

 

Problem 91.      Which one of the following statement does not hold good when two balls of masses m1

elastic collision

  • When m1< m2 and m2 at rest, there will be maximum transfer of momentum

and m2

undergo

 

  • When m1> m2 and m2 at rest, after collision the ball of mass m2 moves with four times the velocity of m1
  • When m1 = m2 and m2 at rest, there will be maximum transfer of kinetic energy
  • When collision is oblique and m2 at rest with m1 = m2 , after collision the balls move in opposite directions

Solution : (b, d) We know that transfer of momentum will be maximum when target is massive and transfer of kinetic energy will be maximum when target and projectile are having same mass. It means statement (a) and (c) are correct, but statement (b) and (d) are incorrect because when target is very light, then after collision it will move with double the velocity of projectile and when collision is oblique and m2 at rest with m1 = m2 , after collision the

ball move perpendicular to each other.

 

 

Perfectly Elastic Oblique Collision.

 

Let two bodies moving as shown in figure. By law of conservation of momentum

Along x-axis, m1u1 + m2u2 = m1v1 cosq + m2v2 cosf

 

 

 

…..(i)

 

Along y-axis, 0 = m v sinqm v

sinf

…..(ii)

 

1 1                   2 2

By law of conservation of kinetic energy

 

 

 

 

1 m u2 + 1 m u2 = 1 m v 2 + 1 m v 2

 

…..(iii)

 

2    1   1      2    2   2       2    1 1      2    2 2

In case of oblique collision it becomes difficult to solve problem when some experimental data are provided as in these situations more unknown variables are involved than equations formed.

 

Special condition : If m1  = m2

u1 = v1 cosq + v2 cosf

0 = v1 sinqv2 sinf

and u2 = 0

substituting these values in equation (i), (ii) and (iii) we get

…..(iv)

…..(v)

 

 

 

and

u2 = v 2 + v 2

…..(vi)

 

1         1        2

Squaring (iv) and (v) and adding we get

 

u2 = v 2 + v 2 + 2v u

cos(q + f)

…..(vii)

 

1         1        2           1 2

Using (vi) and (vii) we get cos(q + f) = 0

\             q + f = p / 2

i.e. after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the

scattering angle q + f would be 90o .

 

Problem 92.      A ball moving with velocity of

9m / s

collides with another similar stationary ball. After the collision both the

 

balls move in directions making an angle of 30o with the initial direction. After the collision their speed will be

 

(a)

2.6m / s

(b)

5.2m / s

(c)

0.52m / s

(d)

52m / s

 

Solution : (b)      Initial horizontal momentum of the system = m ´ 9

Final horizontal momentum of the system = 2mv cos 30o

According to law of conservation of momentum, m ´ 9 = 2mv cos 30o

Þ v = 5.2 m/s

 

 

 

Problem 93.      A ball of mass 1kg , moving with a velocity of 0.4m / s

collides with another stationary ball. After the collision,

 

the first ball moves with a velocity of

  • m / s

in a direction making an angle of

90o

with its initial direction.

 

The momentum of second ball after collision will be (in kg-m/s)

(a) 0.1                             (b) 0.3                          (c) 0.5                             (d) 0.7

Solution : (c)      Let second ball moves with momentum P making an angle q from the horizontal (as shown in the figure).

 

By the conservation of horizontal momentum 1 ´ 0.4 = P cosq By the conservation of vertical momentum     0.3 = P sinq From (i) and (ii) we get P = 0.5 kg-m/s

……(i)

……(ii)

 

 

 

Problem 94.      Keeping the principle of conservation of momentum in mind which of the following collision diagram is not correct

 

 

  • (b) (c)

(d)

 

M1

 

 

M2                               M2

 

 

Solution : (d) In this condition the final resultant momentum makes some angle with x-axis. Which is not possible because initial momentum is along the x-axis and according to law of conservation of momentum initial and final momentum should be equal in magnitude and direction both.

Problem 95. Three particles A, B and C of equal mass are moving with the same velocity v along the medians of an equilateral triangle. These particle collide at the centre G of triangle. After collision A becomes stationary, B retraces its path with velocity v then the magnitude and direction of velocity of C will be

 

  • v and opposite to B
  • v and in the direction of A
  • v and in the direction of C
  • v and in the direction of B

Solution : (d)     From the figure (I) it is clear that before collision initial momentum of the system = 0

After the collision, A becomes stationary, B retraces its path with velocity v. Let C moves with velocity V making an angle q from the horizontal. As the initial momentum of the system is zero, therefore horizontal and vertical momentum after the collision should also be equal to zero.

 

 

From figure (II) Horizontal momentum v cosq + v cos 30 o = 0

Vertical momentum v sinqv sin 30 o = 0

…..(i)

…..(ii)

 

By solving (i) and (ii) we get q = -30o and V = v i.e. the C will move with velocity v in the direction of B.

 

Problem 96.      A ball

B1 of mass M moving northwards with velocity v collides elastically with another ball B2

of same mass

 

but moving eastwards with the same velocity v. Which of the following statements will be true

 

(a)

B1 comes to rest but B2

moves with velocity      2v

 

 

 

(b)

B1 moves with velocity

2v but B2

comes to rest

 

 

(c) Both move with velocity v /           in north east direction

 

(d)

B1 moves eastwards and B2

moves north wards

 

Solution : (d)     Horizontal momentum and vertical momentum both should remain conserve before and after collision. This is possible only for the (d) option.

 

 

 

Head on Inelastic Collision.

  • Velocity after collision : Let two bodies A and B collide inelastically and coefficient of restitution is e.

 

Where

e v2 – v1

u1 – u2

Relative velocity of separation Relative velocity of approach

 

Þ                        v2 – v1 = e(u1 – u2 )

\                         v2 = v1 + e(u1 – u2 )

……(i)

 

From the law of conservation of linear momentum

m1u1 + m2u2 = m1v1 + m2v2

By solving (i) and (ii) we get

……(ii)

 

æ m1em2 ö       æ (1 + e)m2 ö

v1 = ç m   m    ÷ u1 + ç m   m   ÷ u2

è     1          2   ø       è    1         2 ø

é(1 + e)m1 ù        æ m2em1 ö

 

Similarly

v2 = ê m   m   ú u1 + ç m   m    ÷ u2

 

ë    1         2 û        è     1          2 ø

 

By substituting e = 1, we get the value of v1 and u2

for perfectly elastic head on collision.

 

  • Ratio of velocities after inelastic collision : A sphere of mass m moving with velocity u hits inelastically with another stationary sphere of same

 

\                         e v2 – v1

u1 – u2

v2 – v1

u – 0

 

Þ                        v2 – v1 = eu

……(i)

 

By conservation of momentum :

Momentum before collision = Momentum after collision

mu = mv1 + mv2

 

Þ                        v1 + v2 = u

……(ii)

 

Solving equation (i) and (ii) we get v1

= u (1 – e) and v

2                     2

= u (1 + e) 2

 

\                         v1

v2

1 e

1 + e

 

  • Loss in kinetic energy
æ 1 mu2 + 1 mu2ö – æ 1 mv 2 + 1 mv 2ö

÷

ç 21 122   2÷   ç 21 122   2ø

 

Loss (DK) = Total initial kinetic energy – Total final kinetic energy

=

è                            ø    è

Substituting the value of v1 and v2 from the above expression

1 æ  m1m2   ö             2               2

 

Loss (DK) =     ç

2 m + m

÷(1 – e

)(u1 – u2 )

 

è   1         2 ø

By substituting e = 1 we get DK = 0 i.e. for perfectly elastic collision loss of kinetic energy will be zero or kinetic energy remains constant before and after the collision.

 

 

 

 

Problem 97.      A body of mass

40kg

having velocity

4m / s

collides with another body of mass

60kg

having velocity

 

2m / s . If the collision is inelastic, then loss in kinetic energy will be         [CPMT 1996; UP PMT 1996; Pb. PMT 2001]

(a) 440 J                                    (b) 392 J                               (c)  48 J                                     (d) 144 J

 

Solution : (c)      Loss of K.E. in inelastic collision DK = 1         m1m2      (u   u )2 = 1

40 ´ 60

(4 – 2)2 = 1 2400 4 = 48 J.

 

 

2 (m1 + m2 )   1       2

2 (40 + 60)

2 100

 

Problem 98.      One sphere collides with another sphere of same mass at rest inelastically. If the value of coefficient of

 

restitution is

1 , the ratio of their speeds after collision shall be                                                       [RPMT 1998]

2

 

(a) 1 : 2                           (b) 2 : 1                        (c) 1 : 3                           (d) 3 : 1

 

Solution : (c)

v1 = 1 – e = 1 – 1 / 2 = 1 / 2 = 1 .

 

 

v2      1 + e      1 + 1 / 2      3 / 2      3

 

Problem 99. The ratio of masses of two balls is 2 : 1 and before collision the ratio of their velocities is 1 : 2 in mutually opposite direction. After collision each ball moves in an opposite direction to its initial direction. If e = (5/6), the ratio of speed of each ball before and after collision would be

  • (5/6) times (b) Equal

(c) Not related                                                       (d) Double for the first ball and half for the second ball

Solution : (a)      Let masses of the two ball are 2m and m, and their speeds are u and 2u respectively.

By conservation of momentum m1 u1 + m2 u 2 = m1 v1 + m2 v 2 Þ 2mu – 2mu mv2 – 2mv1 Þ v2 = 2v1

 

 

 

 

Coefficient of restitution = – (v 2 – v1 ) = – (2v1 + v1 ) = – 3v1

 

= v1 = 5

[As e = 5

given]

 

(u 2u1 )

(-2uu)

– 3u        u       6                           6

 

Þ v1 u1

= 5 = ratio of the speed of first ball before and after collision.

6

 

Similarly we can calculate the ratio of second ball before and after collision, v2

u2

2v1

2u

v1 u

= 5 .

6

 

Problem 100. Two identical billiard balls are in contact on a table. A third identical ball strikes them symmetrically and come to rest after impact. The coefficient of restitution is

 

(a) 2 3

(b)

1                           (c)

3

1                              (d)     3

6                                       2

 

Solution : (a)

sinq = r   = 1

2r       2

Þ q = 30o

 

From conservation of linear momentum mu = 2mv cos 30 o or

v =  u  

 

Now Relative velocity of separation Relative velocity of approach

in common normal direction.

 

Hence, e =

v          =                  = 2

u cos 30 o                           3

 

Problem 101. A body of mass

3kg , moving with a speed of

4ms-1 , collides head on with a stationary body of mass

2kg .

 

 

 

 

 

Solution: (a,b,c)

Their relative velocity of separation after the collision is 2ms-1 . Then

(a) The coefficient of restitution is 0.5                        (b) The impulse of the collision is 7.2 N-s

  • The loss of kinetic energy due to collision is 6 J (d) The loss of kinetic energy due to collision is 7.2 J

m1 = 3kg , m2 = 2kg , u1 = 4m/s , u2 = 0

Relative velocity of approach u1 – u2 = 4m/s

Relative velocity of separation v2v1 = 2m/s (given)

 

Coefficient of restitution e relative velocity of separation

= 2 = 1 = 0.5

 

 

relative velocity of approach         4     2

 

1   m1m2                 2

2     1 3 ´ 2 é

 

æ 1 ö 2 ù       2

 

Loss in kinetic energy =

2 m + m

(1 – e)

(u1 – u2 )

= 2 3 + 2 ê1 – ç 2 ÷

ú (4)

= 7.2J

 

1         2                                                       êë     è ø úû

 

æ m1 – em2 ö

 

é(1 + e)m2 ù

 

(3 – 0.5 ´ 2)                 8

 

Final velocity of m1 mass,

v1 = ç m   + m

÷ u1 + ê m   + m

ú u2 =

3 + 2

´ 4 + 0 = 5 m/s

 

è      1         2 ø          ë     1         2 û

Impulse of collision = change in momentum of mass m1 (or m2) = m1v1m1u1

= 3 ´ 8 – 3 ´ 4 = 24 – 12 = 4.8 – 12 = -7.2 Ns .

 

5                5

 

Problem 102. Two cars of same mass are moving with same speed v on two different roads inclined at an angle q

with each

 

other, as shown in the figure. At the junction of these roads the two cars collide inelastically and move simultaneously with the same speed. The speed of these cars would be

 

 

 

(a)

 

(b)

 

(c)

v cos q

2

v cos q 2

v cos q

 

2       2

  • 2v cos q

 

Solution : (a)      Initial horizontal momentum of the system = mv cos q +

2

q

mv cos 2 .

 

If after the collision cars move with common velocity V then final horizontal momentum of the system = 2mV.

By the law of conservation of momentum, 2mV = mv cos q + mv cos q Þ V =                          q

2              2            v cos 2 .

 

 

 

Rebounding of Ball After Collision With Ground.

If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.

v0 =                             [From v 2 = u2 + 2gh]

and it rebounds from the floor with a speed

 

 

 

v   = ev  = e

éAs e = velocity after collision ù

 

û

1            0                               ê

ë

velocity before collision ú

v 2          2

 

  • First height of rebound :

\            h1 = e2h0

h1 =   1 = e h0

2g

 

  • Height of the ball after nth rebound : Obviously, the velocity of ball after nth rebound will be

vn  = env0

 

Therefore the height after nth

v 2

rebound will be hn =   n = e

2g

2nh0

 

 

\                         hn = e 2nh0

(3) Total distance travelled by the ball before it stops bouncing

H = h0 + 2h1 + 2h2 + 2h3 + …… = h0 + 2e 2h0 + 2e 4 h0 + 2e 6h0 + …..

 

 

H = h0 [1 + 2e

2 (1 + e 2

  • e 4
  • e 6

….)]

é

= h0 ê1 + 2e

2 æ  1    öù

ç 1 – e 2 ÷ú

éAs

ê

ë

1 + e 2

  • e 4

+ …. =  1    ù

2 ú

1 – e

 

 

 

\            H = h

é1 + e 2 ù

 

ë           è           øû                                                       û

 

0 ê1 – e 2 ú

ë          û

(4) Total time taken by the ball to stop bouncing

 

 

T = t0

+ 2t1

+ 2t 2

+ 2t3

+ …. =

+ 2          + 2

+ ….

 

 

 

 

=             [1 + 2e + 2e 2 +…… ]

[As h = e 2h ; h = e 4 h ]

 

1              0       2              0

 

 

=             [1 + 2e(1 + e + e 2 + e 3 +….. )] =

é + 2eæ   1   öù =

ç        ÷ú
ê1

1 – e

æ 1 + e ö

ç 1 – e ÷

 

 

T =
\

æ 1 + e ö

ç 1 – e ÷

ë         è         øû              è         ø

 

è        ø

 

 

 

 

Problem 103. The change of momentum in each ball of mass

60gm , moving in opposite directions with speeds

4m / s

 

collide and rebound with the same speed, is                                                                                   [AFMC 2001]

 

(a)

0.98 kgm / s

(b)

0.73 kgm / s

(c)

0.48 kgm / s

(d)

0.22kgm / s

 

Solution : (c)      Momentum before collision = mv,          Momentum after collision = – mv

\ Change in momentum = 2mv = 2 ´ 60 ´ 10 -3 ´ 4 = 480 ´ 10 -3 kgm/s = 0.48 kg m/s

Problem 104. A body falling from a height of 20m rebounds from hard floor. If it loses 20% energy in the impact, then coefficient of restitution is                                                                                                        [AIIMS 2000]

(a) 0.89                           (b) 0.56                        (c) 0.23                           (d) 0.18

Solution : (a)      It loses 20% energy in impact and only 80% energy remains with the ball

So ball will rise upto height h2 = 80% of h1 = 80 ´ 20 = 16 m

100

 

Now coefficient of restitution e =                 =            =             = 0.89.

 

Problem 105. A rubber ball is dropped from a height of 5m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8m . The ball loses its velocity on bouncing by a factor of                                            [CBSE PMT 1998]

(a) 16/25                          (b) 2/5                          (c) 3/5                             (d) 9/25

Solution : (c)      If ball falls from height h1, then it collides with ground with speed v1 =                                             …..(i) and if it rebound with velocity v2, then it goes upto height h2 from ground, v2 =                             …..(ii)

 

From (i) and (ii) v2 =

v1

=            =             =

= 3 .

5

 

 

 

Perfectly Inelastic Collision.

In such types of collisions the bodies move independently before collision but after collision as a one single body.

  • When the colliding bodies are moving in the same direction

By the law of conservation of momentum

m1u1 + m2u2 = (m1 + m2 )vcomb

Þ            v          m1u1 + m2u2

 

comb

m1 + m2

 

 

 

Loss in kinetic energy DK = æ 1 m u2 + 1 m  u2 ö – 1 (m

+ m )v 2

 

2

ç

è

1 æ m1m2   ö

 

 

1   1      2

 

2

2   2 ÷          1

ø    2

comb

 

DK =     ç

÷(u1 – u2 )

[By substituting the value of vcomb]

 

2 è m1 + m2 ø

 

 

 

 

 

(2) When the colliding bodies are moving in the opposite direction

By the law of conservation of momentum

 

m u + m (-u ) = (m + m )v

(Taking left to right as positive)

 

1   1         2        2            1          2  comb

v          m1u1 – m2u2

 

comb

m1 + m2

 

when m1u1 > m2u2 then vcomb > 0

(positive)

 

i.e. the combined body will move along the direction of motion of mass m1 .

 

when m1u1 < m2u2 then vcomb < 0

(negative)

 

i.e. the combined body will move in a direction opposite to the motion of mass m1 .

(3) Loss in kinetic energy

DK = Initial kinetic energy – Final kinetic energy

= æ 1 m u2 + 1 m u2 ö – æ 1 (m   + m )v 2     ö

 

2
2

ç      1 1

è

2 2 ÷   ç

2          ø   è

1          2     comb ÷

ø

 

1

= 1   m1m2    (u

2

+ u )2

 

2 m1 + m2

Problem 106. Which of the following is not a perfectly inelastic collision                                        [BHU 1998; JIPMER 2001, 2002]

  • Striking of two glass balls (b) A bullet striking a bag of sand

(c) An electron captured by a proton                            (d) A man jumping onto a moving cart

Solution : (a)      For perfectly elastic collision relative velocity of separation should be zero i.e. the colliding body should move together with common velocity.

 

Problem 107. A metal ball of mass 2kg

moving with a velocity of 36km / h has an head-on collision with a stationary ball of

 

mass 3kg . If after the collision, the two balls move together, the loss in kinetic energy due to collision is

[CBSE 1997; AIIMS 2001]

 

(a)

40 J

(b)

60 J

(c)

100 J

(d)

140 J

 

Solution : (b)      Loss in kinetic energy DK = 1 m1m2  (u u )2 = 1 2 ´ 3 (10 – 0)2 = 60 J.

2 m1 + m2      1        2           2 2 + 3

 

Problem 108. A mass of

20kg

moving with a speed of 10m / s

collides with another stationary mass of 5 kg. As a result of

 

the collision, the two masses stick together. The kinetic energy of the composite mass will be              [MP PMT 2000]

(a) 600 J                                   (b) 800 J                               (c) 1000 J                                 (d) 1200 J

 

Solution : (b)      By conservation of momentum

m1u1 + m2u2 = (m1 + m2 )V

 

Velocity of composite mass V = m1u1 + m2u2

m1 + m2

20 ´ 10 + 5 ´ 0

20 + 5

= 8 m/s

 

\ Kinetic energy of composite mass = 1 (m1 + m2 )V 2 = 1 (20 + 5) ´ 8 2 = 800J.

2                        2

 

 

 

 

Problem 109. A neutron having mass of 1.67 ´ 1027 kg

and moving at 108 m / s

collides with a deutron at rest and sticks to

 

  1. If the mass of the deutron is 3.34 ´ 10-27 kg ; the speed of the combination is [CBSE PMT 2000]

 

(a)

2.56 ´ 103 m / s

(b)

2.98 ´ 105 m / s

(c)

3.33 ´ 107 m / s

(d)

5.01´ 109 m / s

 

Solution : (c)

m1 = 1.67 ´ 10 27 kg , u1 = 108 m/s , m2 = 3.34 ´ 10 27 kg and u2 = 0

 

Speed of the combination V = m1u1 + m2u2

m1 + m2

=       1.67 ´ 10 -27 ´ 108 + 0

1.67 ´ 10 -27 + 3.34 ´ 10 -27

= 3.33 ´ 107 m/s.

 

Problem 110. A particle of mass m moving eastward with a speed v collides with another particle of the same mass moving

northward with the same speed v . The two particles coalesce on collision. The new particle of mass 2m will

move in the north-easterly direction with a velocity                [NCERT 1980; CPMT 1991; MP PET 1999; DPMT 1999]

 

(a)

v / 2

  • 2v
  • v /
  • v

 

Solution : (c)      Initially both the particles are moving perpendicular to each other with momentum mv.

 

So the net initial momentum =                                    =     2 mv .

After the inelastic collision both the particles (system) moves with velocity V, so linear momentum = 2mV

By the law of conservation of momentum          2 mv = 2 mV

 

\ = v/ 2.

 

Problem 111. A particle of mass

m

moving with velocity

v

collides inelastically with a stationary particle of mass

‘2m‘ .

 

The speed of the system after collision will be                                                                               [AIIMS 1999]

 

 

(a)

v                                         (b) 2v

2

(c)

v                                         (d) 3v

3

 

Solution : (c)      By the conservation of momentum mv + 2m ´ 0 = 3 mV

\ V = v .

3

 

Problem 112. A ball moving with speed v hits another identical ball at rest. The two balls stick together after collision. If specific heat of the material of the balls is S, the temperature rise resulting from the collision is                               [Roorkee 1999]

 

  • v2

8S

v2                                   (c)

4S

v2                                       (d)   v2

2S                                                S

 

 

Solution : (b)      Kinetic energy of ball will raise the temperature of the system

1 mv 2 = (2m) S Dt

2

Þ Dt = v2 .

4S

 

Problem 113. A bullet of mass a is fired with velocity b in a large block of mass c. The final velocity of the system will be

 

 

(a)

    c a + c

(b)

  ab a + c

(a + b)

c

(a + c) b a

 

Solution : (b)      Initially bullet moves with velocity b and after collision bullet get embedded in block and both move together with common velocity.

By the conservation of momentum a ´ b + 0 = (a + c) V

 

 

\ =

ab a + c

 

 

 

 

Problem 114.  A particle of mass 1g having velocity  3ˆi – 2ˆj

as  4ˆj – 6kˆ. Velocity of the formed particle is

has a glued impact with another particle of mass 2g and velocity

 

(a)

5.6ms1

(b) 0                             (c)

6.4ms1

(d)

  • ms1

 

Solution : (d)     By conservation of momentum

r

mu1

r

  • m2u2

= (m1

r

  • m2)V

 

r          r           r                       ˆ                ˆ                       ˆ

 

V m1u1 + m2u2

m1 + m2

= 1(3i – 2 j) + 2(4 j – 6k)

m1 + m2

= 3i + 6 j – 12k

(1 + 2)

= ˆi  + 2ˆj – 4kˆ

 

r

=                                        =                         = 4.6ms 1 .

| V |

Problem 115.  A body of mass 2kg is placed on a horizontal frictionless surface. It is connected to one end of a spring whose

force constant is 250 N / m . The other end of the spring is joined with the wall. A particle of mass 0.15kg

moving horizontally with speed v sticks to the body after collision. If it compresses the spring by 10cm , the velocity of the particle is

 

(a)

3m / s

(b)

5m / s

(c)

10m / s

(d)

15m / s

 

Solution : (d)     By the conservation of momentum

Initial momentum of particle = Final momentum of system Þ m ´ v = (m + M) V

 

 

\ velocity of system V =

mv

 

(m + M)

 

Now the spring compresses due to kinetic energy of the system so by the conservation of energy

 

1     2     1

 

 

2     1            æ   mv    ö 2

 

2 kx    = 2 (m + M)V

= 2 (m Mm M ÷

 

 

Þ kx 2 =

m2v 2

m + M

Þ v =

è              ø

 

= x m

 

Putting m = 0.15 kg, M = 2 kg, k = 250 N/m, x = 0.1 m we get v = 15 m/s.

 

 

Collision Between Bullet and Vertically Suspended Block.

A bullet of mass m is fired horizontally with velocity u in block of mass M suspended by vertical thread.

After the collision bullet gets embedded in block. Let the combined system raised upto height h and the string makes an angle q with the vertical.

  • Velocity of system

Let v be the velocity of the system (block + bullet) just after the collision.

Momentumbullet + Momentumblock = Momentumbullet and block system

mu + 0 = (m + M)v

 

\                                                  v     mu    

(m + M)

……(i)

 

 

 

  • Velocity of bullet : Due to energy which remains in the bullet block system, just after the collision, the system (bullet + block) rises upto height h.

 

By the conservation of mechanical energy

1 (m + M)v 2 = (m + M)gh Þ v =

2

 

Now substituting this value in the equation (i) we get

=    mu m + M

 

\                                                                  u = é(m + M) 2gh ù

ë
û

êê        m             ú

  • Loss in kinetic energy : We know the formula for loss of kinetic energy in perfectly inelastic collision

 

1

DK = 1   m1m2    (u

+ u )2

 

2 m1 + m2

 

2

\                         DK = 1

mM   u2

[As u = u , u

= 0 , m   = m and m   = M ]

 

2 m + M

(4) Angle of string from the vertical

1

 

 

é(m + M)

2

 

 

2gh ù

1

 

 

 

u2 æ

2

 

 

 

m      ö 2

 

From the expression of velocity of bullet u = ê                        ú we can get h =      ç           ÷

 

 

 

 

From the figure

ëê        m            úû

q L  h =    – h              u 2 æ    m     ö 2

 

2g è m + M ø

 

cos

1        = 1 – 2      ç              ÷

 

L                 L                gL è m + M ø

é        1 æ   mu   ö 2 ù

or                           q = cos 1 ê1 –         ç            ÷ ú

ëê    2gL è m + M ø úû

 

 

Problem 116. A bullet of mass m

moving with velocity v strikes a block of mass M at rest and gets embeded into it. The

 

kinetic energy of the composite block will be                                                                             [MP PET 2002]

 

 

(a)

1 mv2 ´       m

(b)

1 mv2 ´       M

(c)

1 mv2 ´ (M + m)

 

(d)

1 Mv2 ´       m

 

 

2          (m + M)

2          (m + M)              2              M

2          (m + M)

 

Solution : (a)      By conservation of momentum,

Momentum of the bullet (mv) = momentum of the composite block (m + M)V

Þ Velocity of composite block V = mv

m + M

 

1               2     1

 

 

æ   mv   ö2

 

1  m2v2

1      2æ      m      ö

 

\ Kinetic energy =       (m + M)V

2

= 2 (m M) ç m M ÷

= 2 m + M = 2 mv ç m + M ÷.

 

è              ø                                          è              ø

 

Problem 117. A mass of

10gm , moving horizontally with a velocity of

100cm / sec , strikes the bob of a pendulum and

 

strikes to it. The mass of the bob is also 10gm (see fig.) The maximum height to which the system can be

 

raised is ( g = 10m / sec 2 )

[MP PET 1993; RPMT 1997]

 

 

  • Zero
  • 5cm
  • 5cm
  • 25cm

 

 

 

 

Solution : (d)     By the conservation of momentum,

 

Momentum of the bullet = Momentum of system Þ 10 ´ 1 = (10 + 10) ´ v

Þ v = 1 m/s

2

 

 

Now maximum height reached by system H

 

max

v 2

2g

= (1 / 2)2 m = 1.25cm .

2 ´ 10

 

Problem 118. A bullet of mass m moving with a velocity v strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height h the initial velocity of the bullet is                       [MP PMT 1997]

 

(a)

 

(b)

 

(c)

 

(d)

 

Solution : (a)      By the conservation of momentum mv = (m + M)V

and if the system goes upto height h then V =

 

\ mv = (m + M)             Þ v =                           .

 

Problem 119. A bag P (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (beg + bullet) system the                                                      [CPMT 1989]

 

  • Momentum is mvM 

M + m

  • Kinetic energy

mV 2

 

2

 

 

  • Momentum is

mv (M + m)

M

  • Kinetic energy is

m2V 2

 

2(M + m)

 

Solution : (d)     Velocity of combined system V = mv 

m + M

Momentum for combined system = (m + M)V = (m + M) mv 

m + M

 

1               2    1

 

 

æ   mv    ö2     1

 

 

m2v2

m2v2

 

Kinetic energy for combined system =         (m + M)V

2

= 2 (m M) ç m M ÷

= 2 (m M) (m + M)2 = 2(m M) .

 

è              ø

Problem 120. A wooden block of mass M is suspended by a cord and is at rest. A bullet of mass m, moving with a velocity v pierces through the block and comes out with a velocity v / 2 in the same direction. If there is no loss in kinetic energy, then upto what height the block will rise

 

(a)

m2v2 / 2M 2g

(b)

m2v2 / 8M 2g

(c)

m2v2 / 4 Mg

(d) m2v2 / 2Mg

 

Solution : (b)      By the conservation of momentum

Initial momentum = Final momentum

 

mv + M ´ 0 = m v + M ´ V

2

Þ V =  m  v

2M

 

If block rises upto height h then h = V 2

2g

 

= (mv / 2M)2

2g

 

m2v 2   .

8M 2 g

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