Center of Mass and Collision — Page 2 | NEET Physics

Q11

Body A of mass 4m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :

A

{5 \over 9}

B

{4 \over 9}

C

{8 \over 9}

D

{1 \over 9}

Correct Answer

Option C

Solution

As linear momentum is conserved before and after the collision, $\therefore$ 4m $\times$ u = 4m $\times$ v 1 + 2m $\times$ v 2 $\Rightarrow$ 2u = 2v 1 + v 2 ......(

1) As collision is elastic so e = 1 We know, e =

{{{v_2} - {v_1}} \over {{u_1} - {u_2}}}

$\Rightarrow$ 1 =

{{{v_2} - {v_1}} \over {u - 0}}

$\Rightarrow$ u = v 2 - v 1 ....(2) Subtractiong (2) from (1), we get u = 3v 1 $\Rightarrow$ v 1 =

{u \over 3}

After the collision the fraction of energy lost by the colliding body A is =

{{{1 \over 2}\left( {4m} \right){u^2} - {1 \over 2}\left( {4m} \right){{\left( {{u \over 3}} \right)}^2}} \over {{1 \over 2}\left( {4m} \right){u^2}}}

=

{{1 - {1 \over 9}} \over 1}

=

{8 \over 9}

Q12

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

A 0.5

B 0.25

C 0.8

D 0.4

Correct Answer

Option B

Solution

From the law of conservation of linear momentum, momentum before collision = momentum after collision, so mv + 4m × 0 = m × 0 + 4mv′ mv = 4mv′ or v = 4v′ Coefficient of restitution, e =

{{v'} \over v} = {{{v \over 4}} \over v}

= 0.25

Q13

Which of the following statements are correct ? (1) Centre of mass of a body always coincides with the centre of gravity of the body. (2) Centre of mass of a body is the point at which the total gravitational torque on the body is zero. (3) A couple on a body produces both translational and rotational motion in a body (4) Mechanical advantage greater than one means that small effort can be used to lift a large load.

A (1) and (2)

B (2) and (3)

C (3) and (4)

D (4)

Correct Answer

Option D

Solution

Centre of gravity of a body is the point at which the total gravitational torque on body is zero.

Centre of mass and centre of gravity coincides only for symmetrical bodies.

Hence statements (1) and (2) are incorrect.

A couple of a body produces rotational motion only.

Hence statement (3) is incorrect.

Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it.

Hence statement (4) is correct.

Q14

Two identical balls A and B having velocities of 0.5 m s

-

1 and

-

0.3 m s

-

1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be :

A

-

0.5 m s

-

1 and 0.3 m s

-

1

B 0.5 m s

-

1 and

-

0.3 m s

-

1

C

-

0.3 m s

-

1 and 0.5 m s

-

1

D 0.3 m s

-

1 and 0.5 m s

-

1

Correct Answer

Option B

Solution

Masses of the balls are same and collision is elastic, so their velocity will be interchanged after collision.

Q15

A bullet of mass 10 g moving horizontally with a velocity of 400 m s

-

1 strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

A 100 m s

-

1

B 80 m s

-

1

C 120 m s

-

1

D 160 m s

-

1

Correct Answer

Option C

Solution

Mass of bullet, m = 10 g = 0.01 kg Initial speed of bullet, u = 400 m s –1 Mass of block, M = 2 kg Length of string, l = 5 m Speed of the block after collision = v1 Speed of the bullet on emerging from block, v = ?

Using energy conservation principle for the block, (KE + PE) Reference = (KE + PE) h

\Rightarrow {1 \over 2}Mv_1^2 = Mgh\,or,\,{v_1} = \sqrt {2gh}

{v_1} = \sqrt {2 \times 10 \times 0.1} = \sqrt 2 \,m{s^{ - 1}}

Using momentum conservation principle for block and bullet system, (M × 0 + mu) Before collision = (M × v 1 + mv) After collision

\Rightarrow 0.01 \times 400 = 2\sqrt 2 + 0.01 \times v

\Rightarrow v = {{4 - 2\sqrt 2 } \over {0.01}} = 117.15\,m{s^{ - 1}} \approx 120\,m{s^{ - 1}}

Q16

A rigid ball of mass m strikes a rigid wall at 60 o and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be :

A

mV

B 2

mV

C

{{mV} \over 2}

D

{{mV} \over 3}

Correct Answer

Option A

Solution

Given, p i = p f = mV Change in momentum of the ball

= {\overrightarrow p _f} - {\overrightarrow p _i}

= \left( { - {p_{fx}}\widehat i - {p_{fy}}\widehat j} \right) - \left( { - {p_{ix}}\widehat i - {p_{iy}}\widehat j} \right)

= - \widehat i\left( {{p_{fx}} + {p_{ix}}} \right) - \widehat j\left( {{p_{fy}} + {p_{iy}}} \right)

= - 2{p_{ix}}\widehat i = - mV\widehat i\left[ \because {{p_{fy}} - {p_{iy}} = 0} \right]

Here, p ix = p fx = p i cos 60° =

{{mV} \over 2}

$\because$ Impulse imparted by the wall = change in the momentum of the ball = mV.

Q17

Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is :

A 7.5R

B 1.5R

C 2.5R

D 4.5R

Correct Answer

Option A

Solution

Let the distance moved by spherical body of mass M is x 1 and by spherical body of mass 5m is x 2 .

As their C.M. will remain stationary $\therefore$ (M) (x 1 ) = (5M) (x 2 ) $\Rightarrow$ x 1 = 5x 2 x 1 + x 2 = 9R So, x 1 = 7.5 R

Q18

Two particles of masses m 1 , m 2 move with initial velocities u 1 and u 2 . On collision, one of the particles get excited to higher level, after absorbing energy

\varepsilon

. If final velocities of particles be v 1 and v 2 then we must have :

A

{1 \over 2}

m 1 u

_1^2

+

{1 \over 2}

m 2 u

_2^2

-

\varepsilon

=

{1 \over 2}

m 1 v

_1^2

+

{1 \over 2}

m 2 v

_2^2

B

{1 \over 2}

m

_1^2

u

_1^2

+

{1 \over 2}

m

_2^2

u

_2^2

+

\varepsilon

=

{1 \over 2}

m

_1^2

v

_1^2

+

{1 \over 2}

m

_2^2

v

_2^2

C m

_1^2

u 1 + m

_2^2

u 2

-

\varepsilon

= m

_1^2

v 1 + m

_2^2

v 2

D

{1 \over 2}

m 1 u

_1^2

+

{1 \over 2}

m 2 u

_2^2

=

{1 \over 2}

m 1 v

_1^2

+

{1 \over 2}

m 2 v

_2^2

-

\varepsilon

Correct Answer

Option A

Solution

By law of conservation of energy, K.E f = K.E i – excitation energy (e) $\therefore$

{1 \over 2}m_1v_1^2 + {1 \over 2}m_2v_2^2 = {1 \over 2}{m_1}u_1^2 + {1 \over 2}{m_2}u_2^2 - \varepsilon

Q19

The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is :

A 24 N s

B 20 N s

C 12 N s

D 6 N s

Correct Answer

Option C

Solution

Change in momentum = Area under F-t graph in that interval = Area of

\Delta

ABC – Area of rectangle CDEF + Area of rectangle FGHI =

{1 \over 2}

$\times$ 2 $\times$ 6 - 3 $\times$ 2 + 4 $\times$ 3 = 12 N s

Q20

A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is :

A mv 2

B

{3 \over 2}

mv 2

C 2mv 2

D 4mv 2

Correct Answer

Option B

Solution

By conservation of linear momentum

2m{v_1} = \sqrt 2 mv \Rightarrow {v_1} = {v \over {\sqrt 2 }}

As two masses of each of mass m move perpendicular to each other. Total KE generated

= {1 \over 2}m{v^2} + {1 \over 2}m{v^2} + {1 \over 2}\left( {2m} \right)v_1^2

= m{v^2} + {{m{v^2}} \over 2} = {3 \over 2}m{v^2}