Center of Mass and Collision — Page 5 | NEET Physics

Q41

Particle A of mass m1 moving with velocity

\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}

collides with another particle B of mass m2 which is at rest initially. Let

\overrightarrow {{V_1}}

and

\overrightarrow {{V_2}}

be the velocities of particles A and B after collision respectively. If m1 = 2m2 and after collision

\overrightarrow {{V_1}} =

\left( {\widehat i + \sqrt 3 \widehat j} \right)

, the angle between

\overrightarrow {{V_1}}

and

\overrightarrow {{V_2}}

is :

A 105o

B 15o

C -45o

D 60o

Correct Answer

Option A

Solution

Given m1 = 2m2 So let, m2 = m and m1 = 2m From momentum conservation

{\overrightarrow p _i}

=

{\overrightarrow p _f}

$\Rightarrow$ (2m)

\left( {\sqrt 3 \widehat i + \widehat j} \right)

+ 0 = 2m

\left( {\widehat i + \sqrt 3 \widehat j} \right)

+ m

{\overrightarrow V _2}

$\Rightarrow$

{\overrightarrow V _2}

= 2

\left( {\sqrt 3 \widehat i + \widehat j} \right)

- 2

\left( {\widehat i + \sqrt 3 \widehat j} \right)

$\Rightarrow$

{\overrightarrow V _2}

=

\left( {2\sqrt 3 - 2} \right)\widehat i - \widehat j\left( {2\sqrt 3 - 2} \right)

=

2\left( {\sqrt 3 - 1} \right)\left( {\widehat i - \widehat j} \right)

Also given after collision

\overrightarrow {{V_1}} = \left( {\widehat i + \sqrt3\widehat j} \right)m{s^{ - 1}}

For angle between

{\overrightarrow V _1}

&

{\overrightarrow V _2}

, cos $\theta$ =

{{{{\overrightarrow V }_1}.{{\overrightarrow V }_2}} \over {\left| {{{\overrightarrow V }_1}} \right|\left| {{{\overrightarrow V }_2}} \right|}}

=

{{2\left( {\sqrt 3 - 1} \right) \times 1 - 2\left( {\sqrt 3 - 1} \right) \times \sqrt 3 } \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

=

{{2\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)} \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

=

{{\left( {1 - \sqrt 3 } \right)} \over {2\sqrt 2 }}

$\Rightarrow$ $\theta$ = 105o

Q42

A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the man with respect to the surface is :

A 0.28 ms–1

B 0.47 ms–1

C 0.20 ms–1

D 0.14 ms–1

Correct Answer

Option C

Solution

50 V1 = 20 V2 V1 + V2 = 0.70 V1 = 0.20

Q43

Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :

A 4 cm towards the 10 kg block

B 2 cm away from the 10 kg block

C 2 cm towards the 10 kg block

D 4 cm away from the 10 kg block

Correct Answer

Option C

Solution

For COM to remain unchanged, m1x1 = m2x2 $\Rightarrow$ 10 $\times$ 6 = 30 $\times$ x2 $\Rightarrow$ x2 = 2 cm towards 10 kg block.

Q44

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms

-

1 collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?

A 100 N

B 200 N

C 300 N

D 400 N

Correct Answer

Option B

Solution

Change in momentum of one ball = 2 $\times$ (0.05)(10) kg m/s = 1 kg m/s

\Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}

N = 200 N

Q45

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}

Q46

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

Q47

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.

Q48

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

Q49

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

Q50

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}