Center of Mass and Collision — Page 6 | NEET Physics

Q51

A body of mass M moving at speed V0 collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles

\theta

1 and

\theta

2 with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles

\theta

1 and

\theta

2 will be equal, is :

A 4

B 1

C 3

D 2

Correct Answer

Option C

Solution

Given $\theta$ 1 = $\theta$ 2 = $\theta$ from momentum conservation in x-direction MV0 = MV1 cos $\theta$ + mV2 cos $\theta$ in y-direction 0 = MV1 sin $\theta$ $-$ mV2 sin $\theta$ Solving above equations

{V_2} = {{M{V_1}} \over m}

, V0 = 2V1 cos $\theta$ From energy conservation

{1 \over 2}MV_0^2 = {1 \over 2}MV_1^2 + {1 \over 2}MV_2^2

Substituting value of V2 & V0, we will get

{M \over m} + 1 = 4{\cos ^2}\theta \le 4

{M \over m} \le 3

Option (c)

Q52

Consider the following two statements :

A.

Linear momentum of a system of particles is zero

B.

Kinetic energy of a system of particles is zero. then

A

A

does not imply

B

and

B

does not imply

A

B

A

implies

B

but

B

does not imply

A

C

A

does not imply

B

but

B

implies

A

D

A

implies

B

and

B

implies

A

Correct Answer

Option A

Solution

The correct answer is Option A :

A

does not imply

B

and

B

does not imply

A

. Here's why : Statement

A:

The linear momentum of a system of particles being zero does not mean that the kinetic energy is also zero.

For example, consider two equal mass particles moving with the same speed but in opposite directions.

The linear momentum of the system will be zero because momentum is a vector quantity and the two momenta will cancel out.

However, kinetic energy is a scalar quantity and does not cancel out in this way.

Each particle has kinetic energy due to its motion, so the total kinetic energy of the system is not zero.

Statement

B:

The kinetic energy of a system of particles being zero also does not imply that the linear momentum is zero.

If the kinetic energy is zero, it means that all the particles are at rest (since kinetic energy is associated with motion).

However, the linear momentum will also be zero in this case because momentum depends on both mass and velocity, and the velocity of each particle is zero.

But remember that zero linear momentum doesn't exclusively mean all particles are at rest.

As explained above, it could be the scenario where particles have equal and opposite momenta, thereby cancelling each other.

Hence,

B

doesn't imply

A

. So, neither statement implies the other. Hence, Option A is correct.

Q53

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

A

{{{v_0}} \over {\sqrt 2 }}

B

{{v_0}} \over 4

C

\sqrt 2 {v_0}

D

{{v_0}} \over 2

Correct Answer

Option C

Solution

From conservation of linear momentum, mv0 = mv1 + mv2 or v0 = v1 + v2 ........(1) According to the question, Kf =

{3 \over 2}

Ki $\Rightarrow$

{1 \over 2}mv_1^2 + {1 \over 2}mv_2^2 = {3 \over 2} \times {1 \over 2}mv_0^2

$\Rightarrow$

v_1^2 + v_2^2 = {3 \over 2}v_0^2

Using eq (1)

{\left( {{v_1} + {v_2}} \right)^2} = v_0^2

$\Rightarrow$

v_1^2 + v_2^2 + 2{v_1}{v_2}

=

v_0^2

$\Rightarrow$

2{v_1}{v_2}

=

v_0^2 - {3 \over 2}v_0^2

=

- {1 \over 2}v_0^2

Now,

{\left( {{v_1} - {v_2}} \right)^2}

=

{\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2}

=

v_0^2 - \left( { - v_0^2} \right)

=

2v_0^2

$\therefore$

{{v_1} - {v_2}}

=

\sqrt 2 {v_0}

Q54

A block of mass

0.50

kg

is moving with a speed of

2.00

m{s^{ - 1}}

on a smooth surface. It strike another mass of

1.0

kg

and then they move together as a single body. The energy loss during the collision is :

A

0.16J

B

1.00J

C

0.67J

D

0.34

J

Correct Answer

Option C

Solution

Let

m

= 0.50 kg and

M

= 1.0 kg Initial kinetic energy of the system when 1 kg mass is at rest,

K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}

= {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J

For collision, applying conservation of linear momentum

\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v

$\therefore$

0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s

Final kinetic energy of the system is

K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}

= {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J

$\therefore$ Energy loss during collision

= \left( {1 - {1 \over 3}} \right)J = 0.67J

Q55

Consider a two particle system with particles having masses

{m_1}

and

{m_2}

. If the first particle is pushed towards the center of mass through a distance

d,

by what distance should the second particle is moved, so as to keep the center of mass at the same position?

A

{{{m_2}} \over {{m_1}}}\,\,d

B

{{{m_1}} \over {{m_1} + {m_2}}}d

C

{{{m_1}} \over {{m_2}}}d

D

d

Correct Answer

Option C

Solution

Initially,

0 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}

Finally, mass m1 moved towards the center a distance d so the distance of mass m1 from the origin is x1 - d, and now let mass m2 need to move d' to keep the center at the origin.

$\therefore$

0 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}

\Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'

\Rightarrow d' = {{{m_1}} \over {{m_2}}}d

Q56

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90o with respect to each other. The mass of unknown particle is :

A

{m \over 2}

B m

C

{m \over {\sqrt 3 }}

D 2 m

Correct Answer

Option B

Solution

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v1 is initial velocity. (i) Before collision : Now, in figure (ii), v1 is final velocity of unknown particle and v2 is final velocity of proton. (ii) After collision : By conservation of momentum, we have Momentum before collision = Momentum after collision Consider x-component, we have

m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ

m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})

..... (1) Consider y-component, we have

0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ

{1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}

...... (2) Substitute Eq. (2) in Eq. (1), we get

m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}

\Rightarrow {v_i} = \sqrt 2 {v_2}

..... (3) Using Eq. (2) and (3) in Eq. (1), we get

m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})

2m{v_2} = 2m'{v_1}

( $\because$

{v_1} = {v_2}

)

\Rightarrow m = m'

Q57

A particle of mass

m

moving in the

x

direction with speed

2v

is hit by another particle of mass

2m

moving in the

y

direction with speed

v.

If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

A

56\%

B

62\%

C

44\%

D

50\%

Correct Answer

Option A

Solution

Applying conservation of linear momentum in x direction

2mv = \left( {2m + m} \right){V_x}

\Rightarrow {V_x} = {2 \over 3}v

Applying conservation of linear momentum in y direction

2mv = \left( {2m + m} \right){V_y}

\Rightarrow {V_y} = {2 \over 3}v

Final speed of 3m mass,

{V_f}

=

\sqrt {V_x^2 + V_y^2}

=

\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}}

=

\sqrt {{{8{v^2}} \over 9}}

Initial kinetic energy

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

= 2m{v^2} + m{v^2}

=

3m{v^2}

Final kinetic energy,

{E_f} = {1 \over 2}\left( {3m} \right)

{V_f^2}

= {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]

= {{4m{v^2}} \over 3}

Energy loss =

{E_i} - {E_f}

=

3m{v^2} - {{4m{v^2}} \over 3}

=

{{5m{v^2}} \over 3}

Percentage loss in the energy during the collision =

{{{E_i} - {E_f}} \over {{E_i}}}

\times 100

=

{{{{5m{v^2}} \over 3}} \over {3m{v^2}}}

\times 100

=

{5 \over 9} \times 100

\simeq 56\%

Q58

A body of mass 2 kg makes an eleastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?

A 1.2 kg

B 1.0 kg

C 1.8 kg

D 1.5 kg

Correct Answer

Option A

Solution

By conservation of linear momentum:

2{v_0} = 2\left( {{{{v_0}} \over 4}} \right) + mv \Rightarrow 2{v_0} = {{{v_0}} \over 2} + mv

\Rightarrow {{3{v_0}} \over 2} = mv\,\,...(1)

Since collision is elastic

{V_{separation}} = {V_{approch}}

\Rightarrow v - {{{v_0}} \over 4} = {v_0} \Rightarrow m = {6 \over 5} = 1.2\,kg

Q59

A simple pendulum, made of a string of length

\ell

and a bob of mass m, is released from a small angle

{{\theta _0}}

. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle

{{\theta _1}}

. Then M is given by :

A

{m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)

B

{m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)

C

m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)

D

m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)

Correct Answer

Option C

Solution

v =

\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)}

v1 =

\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)}

By momentum conservation m

\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}

= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)}

$\Rightarrow$

m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}

$=$ MVm and e = 1 =

{{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}

\sqrt {2g\ell }

\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right)

$=$ Vm . . .(I) m

\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)

$=$ MVM . . .(II) Dividing

{{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}

By componendo divided

{{m - M} \over {m + M}}

=

{{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}

$\Rightarrow$

{M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}

Q60

A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms

-

1 and 40 ms

-

1 respectively. The velocity of the third piece will be :

A 15 ms

-

1

B 25 ms

-

1

C 35 ms

-

1

D 50 ms

-

1

Correct Answer

Option B

Solution

Given problem a body of mass $M$ explodes into three pieces of mass ratio $1: 1: 2$ $\therefore$ Mass of fragments will be $x, x, 2 x$ Hence, $M=x+x+2 x=4 x \mathrm{~kg}$ As in the process of explosion no external forces are involved and explosion occurs due to internal forces.

Thus, momentum of the system will be conserved.

$p_{\text{initial }}=p_{\text{final }}$ By law of conservation of momentum,

M \times 0=\frac{M}{4} \times 30 \hat{i}+\frac{M}{4} \times 40 \hat{j}+\frac{2 M}{4} \vec{v}

Where $\vec{v}$ is the velocity of the third fragment.

\frac{M \vec{v}}{2} =-\frac{M}{4}(30 \hat{i}+40 \hat{j})

$\Rightarrow$

\vec{v} =-15 \hat{i}-20 \hat{j}

Thus, magnitude of $\vec{v}=|\vec{v}|=\sqrt{v_x^2+v_y^2}$ $=\sqrt{(-15)^2+(-20)^2}$

|\vec{v}|=\sqrt{625}=25 \mathrm{~m} / \mathrm{s}