Center of Mass and Collision — Page 2 | JEE Physics

Q11

A body of mass

1000 \mathrm{~kg}

is moving horizontally with a velocity

6 \mathrm{~m} / \mathrm{s}

. If

200 \mathrm{~kg}

extra mass is added, the final velocity (in

\mathrm{m} / \mathrm{s}

) is:

A 6

B 2

C 3

D 5

Correct Answer

Option D

Solution

Momentum will remain conserve

\begin{aligned} & 1000 \times 6=1200 \times v \\ & v=5 \mathrm{~m} / \mathrm{s} \end{aligned}

Q12

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head-on elastic collision)

A 50.0%

B 66.6%

C 55.6%

D 33.3%

Correct Answer

Option C

Solution

For a head on elastic collision

{v_2} = {{m{u_1}} \over {m + 5m}} + {{m{u_1}} \over {m + 5m}}

= {{2{u_1}} \over 6}

or

{{{u_1}} \over 3}

Initial kinetic energy of first mass

= {1 \over 2}mu_1^2

Final kinetic energy of second mass

= {1 \over 2} \times 5m{\left( {{{{u_1}} \over 3}} \right)^2}

= {5 \over 9}\left( {{1 \over 2}mu_1^2} \right)

$\Rightarrow$ kinetic energy transferred = 55% of initial kinetic energy of first colliding mass

Q13

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision. Reason R : During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below :

A A is correct but R is not correct.

B A is not correct but R is correct.

C Both A and R are correct and R is the correct explanation of A.

D Both A and R are correct but R is NOT the correct explanation of A.

Correct Answer

Option C

Solution

m < < M e =

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

For elastic collision $\to$ e = 1 1 =

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

u = v2 $-$ u v2 = 2u In elastic collision kinetic energy & momentum are conserved.

Q14

A body of mass

10 \mathrm{~kg}

is projected at an angle of

45^{\circ}

with the horizontal. The trajectory of the body is observed to pass through a point

(20,10)

. If

\mathrm{T}

is the time of flight, then its momentum vector, at time

\mathrm{t}=\dfrac{\mathrm{T}}{\sqrt{2}}

, is _____________. [Take

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}

]

A

100 \hat{i}+(100 \sqrt{2}-200) \hat{j}

B

100 \sqrt{2} \hat{i}+(100-200 \sqrt{2}) \hat{j}

C

100 \hat{i}+(100-200 \sqrt{2}) \hat{j}

D

100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}

Correct Answer

Option D

Solution

m = 10 kg $\theta$ = 45

^\circ

y = x\tan \theta \left( {1 - {x \over R}} \right)

\Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)

\Rightarrow R = 40

40 = {{{u^2}} \over {10}} \Rightarrow u = 20

\Rightarrow T = {{20 \times 20 \times {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s

at

t = 2,\,\overrightarrow v = \left( {10\sqrt 2 \widehat i} \right) + \left( {10\sqrt 2 - 2 \times 10} \right)\widehat j

\Rightarrow \overrightarrow p = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2 - 20} \right)\widehat j} \right]

= 100\sqrt 2 \widehat i + \left( {100\sqrt 2 - 200} \right)\widehat j

Q15

A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :

A 10.2 eV

B 16.8 eV

C 12.1 eV

D 20.4 eV

Correct Answer

Option D

Solution

Let, velocity offer collision = v1 $\therefore$ From conservation of momentum, mv = (m + m) v1 $\Rightarrow$ v1 =

{v \over 2}

$\therefore$ Loss in kinetic energy =

{1 \over 2}

mv2 $-$

{1 \over 2}

(2m) $\times$

{\left( {{v \over 2}} \right)^2}

=

{1 \over 4}\,

mv2 lost kinetic energy is used by the electron to jump from first orbit to second orbit. $\therefore$

{1 \over 4}

mv2 = (13.6 $-$ 3.4) eV = 10.2 eV $\Rightarrow$

{1 \over 2}

mv2 = 20.4 eV

Q16

A stationary particle breaks into two parts of masses

m_A

and

m_B

which move with velocities

v_A

and

v_B

respectively. The ratio of their kinetic energies

\left(K_B: K_A\right)

is :

A

v_B: v_A

B

1: 1

C

m_B v_B: m_A v_A

D

m_B: m_A

Correct Answer

Option A

Solution

A stationary particle breaks into two parts with masses

m_A

and

m_B

, which then move with velocities

v_A

and

v_B

, respectively. We need to determine the ratio of their kinetic energies

K_A

and

K_B

.

Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:

m_A v_A = m_B v_B

Here, the ratio of kinetic energies is given by:

\frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2}

However, using the momentum relationship, we can substitute

m_A v_A = m_B v_B

into the kinetic energy ratio, leading us to simplify:

\frac{K_A}{K_B} = \frac{v_A}{v_B}

Therefore, the ratio of their kinetic energies

\left(K_B: K_A\right)

is:

\frac{K_B}{K_A} = \frac{v_B}{v_A}

Q17

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)

Q18

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.

Q19

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}

Q20

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