Center of Mass and Collision — Page 9 | NEET Physics

Q81

A ball of mass

0.15 \mathrm{~kg}

hits the wall with its initial speed of

12 \mathrm{~ms}^{-1}

and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is

100 \mathrm{~N}

, calculate the time duration of the contact of ball with the wall.

A 0.018 s

B 0.036 s

C 0.009 s

D 0.072 s

Correct Answer

Option B

Solution

F = 100 N

\Delta

P = 2 $\times$ 0.15 $\times$ 12 = 3.6 $\Rightarrow$ t =

{{3.6} \over {100}}

= 0.036 s

Q82

A circular disc of radius

R

is removed from a bigger circular disc of radius

2R

such that the circumferences of the discs coincide. The center of mass of the new disc is

\alpha R

form the center of the bigger disc. The value of

\alpha

is

A

1/4

B

1/3

C

1/2

D

1/6

Correct Answer

Option B

Solution

Let the mass per unit area be

\sigma .

Then the mass of the complete disc

= \sigma \left[ {\pi {{\left( {2R} \right)}^2}} \right] = 4\pi \sigma {R^2}

The mass of the removed disc

= \sigma \left( {\pi {R^2}} \right) = \pi \sigma {R^2}

So mass of the remaining disc =

4\pi \sigma {R^2}

-

\pi \sigma {R^2}

=

3\pi \sigma {R^2}

Let center of mass of

3\pi \sigma {R^2}

mass is at x distance from origin O. $\therefore$

{{3\pi {R^2}\sigma .x + \pi {R^2}\sigma .R} \over {4\pi {R^2}\sigma }} = 0

As center of mass of full disc is at Origin. $\therefore$

x = - {R \over 3}

According to the question,

x

=

\alpha R

$\therefore$

\alpha = - {1 \over 3}

$\Rightarrow$

\left| \alpha \right| = {1 \over 3}

Q83

Two particles of equal mass m have respective initial velocities

u\widehat i

and

u\left( {{{\widehat i + \widehat j} \over 2}} \right)

. They collide completely inelastically. The energy lost in the process is :

A

{1 \over 3}m{u^2}

B

{1 \over 8}m{u^2}

C

{3 \over 4}m{u^2}

D

\sqrt {{2 \over 3}} m{u^2}

Correct Answer

Option B

Solution

\overrightarrow {{P_i}} = \overrightarrow {{P_f}}

$\Rightarrow$ mu

\widehat i

+ m

\left( {{u \over 2}\widehat i + {u \over 2}\widehat j} \right)

= 2m

\overrightarrow v

Compare both side

\overrightarrow v =

{{3u} \over 4}\widehat i + {u \over 4}\widehat j

$\Rightarrow$

{\left| {\overrightarrow v } \right|^2}

=

{{10{u^2}} \over {16}}

\Delta

KE = KEf – KEi =

{1 \over 2}2m \times {{10{u^2}} \over {16}}

-

{1 \over 2}m{u^2}

-

{1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}

=

- {1 \over 8}m{u^2}

Q84

A machine gun of mass

10 \mathrm{~kg}

fires

20 \mathrm{~g}

bullets at the rate of 180 bullets per minute with a speed of

100 \mathrm{~m} \mathrm{~s}^{-1}

each. The recoil velocity of the gun is

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

Momentum of bullets per unit time

= {{180 \times {{20} \over {1000}} \times 100} \over {60}}

kg m/s

^2

= 6 N $\Rightarrow$ Force on gun = 6 N We cannot calculate recoil velocity with the given data.

If we consider recoil velocity at

t = 1

s, then

{V_{\mathrm{recoil}}} = u + at

= 0 + {6 \over {10}} \times 1

= 0.6 m/s

Q85

A bullet of

10 \mathrm{~g}

leaves the barrel of gun with a velocity of

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

. If the barrel of gun is

50 \mathrm{~cm}

long and mass of gun is

3 \mathrm{~kg}

, then value of impulse supplied to the gun will be :

A 12 Ns

B 3 Ns

C 6 Ns

D 36 Ns

Correct Answer

Option C

Solution

First, we need to find the velocity of the gun after the bullet is fired.

We can use conservation of momentum to do this.

The total momentum of the system of the gun and bullet is conserved before and after the bullet is fired.

Therefore, we can write

m_g u_g + m_b u_b = m_g v_g + m_b v_b

where

u_g = 0

is the initial velocity of the gun,

u_b = 600 \mathrm{~m/s}

is the initial velocity of the bullet,

m_g = 3 \mathrm{~kg}

is the mass of the gun,

m_b = 0.01 \mathrm{~kg}

is the mass of the bullet,

v_b = 0

is the final velocity of the bullet (since it has left the gun), and

v_g

is the final velocity of the gun. Substituting the given values, we get

(3 \mathrm{~kg})(0) + (0.01 \mathrm{~kg})(600 \mathrm{~m/s}) = (3 \mathrm{~kg})v_g + (0.01 \mathrm{~kg})(0)

Solving for

v_g

, we get

v_g = \frac{0.01 \mathrm{~kg} \times 600 \mathrm{~m/s}}{3 \mathrm{~kg}} = 2 \mathrm{~m/s}

Therefore, the velocity of the gun after the bullet is fired is

v_g = 2 \mathrm{~m/s}

.

Next, we need to find the impulse on the gun.

The impulse-momentum theorem states that the impulse on an object is equal to the change in momentum of that object.

Therefore, the impulse on the gun is given by

I = \Delta p = m_g \Delta v

where

\Delta v = v_f - u_g

is the change in velocity of the gun. Since the initial velocity of the gun is zero, we can write

\Delta v = v_g

. Substituting the given values, we get

I = (3 \mathrm{~kg}) \times (2 \mathrm{~m/s}) = 6 \mathrm{~Ns}

Therefore, the impulse on the gun is

6 \mathrm{~Ns}

, which is the correct answer.

Q86

This question has statement

{\rm I}

and statement

{\rm I}

{\rm I}

. Of the four choices given after the statements, choose the one that best describes the two statements. Statement -

{\rm I}

: A point particle of mass

m

moving with speed

\upsilon

collides with stationary point particle of mass

M.

If the maximum energy loss possible is given as

f\left( {{1 \over 2}m{v^2}} \right)

, then

f = \left( {{m \over {M + m}}} \right).

Statement -

{\rm II}

: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

A Statement -

{\rm I}

is true, Statement -

{\rm II}

is true; Statement -

{\rm II}

is the correct explanation of Statement -

{\rm I}

.

B Statement -

{\rm I}

is true, Statement -

{\rm II}

is true; Statement -

{\rm II}

is not the correct explanation of Statement -

{\rm I}

.

C Statement -

{\rm I}

is true, Statement -

{\rm II}

is false

D Statement -

{\rm I}

is false, Statement -

{\rm II}

true.

Correct Answer

Option D

Solution

Initial energy =

{{{P^2}} \over {2m}}

, where

P

is the momentum and m is the mass of the moving particle.

Loss of energy is maximum when collision is inelastic means when the particles get stuck together as a result of the collision.

So after collision energy =

{{{P^2}} \over {2\left( {m + M} \right)}}

$\therefore$ Maximum energy loss

= {{{P^2}} \over {2m}} - {{{P^2}} \over {2\left( {m + M} \right)}}

.

\left[ \right.

As

\left. {K.E. = {{{P^2}} \over {2m}} = {1 \over 2}m{v^2}\,\,} \right]

= {{{P^2}} \over {2m}}\left[ {{M \over {\left( {m + M} \right)}}} \right] = {1 \over 2}m{v^2}\left\{ {{M \over {m + M}}} \right\}

$\therefore$

f = \left( {{M \over {m + M}}} \right)

So statement

I

is wrong. Statement

{\rm I}{\rm I}

says "Maximum energy loss occurs when the particles get stuck together as a result of the collision."

This is a case of perfectly inelastic collision.

Hence statement

{\rm I}

{\rm I}

is correct.

Q87

100 balls each of mass

\mathrm{m}

moving with speed

v

simultaneously strike a wall normally and reflected back with same speed, in time

\mathrm{t ~s}

. The total force exerted by the balls on the wall is

A

\dfrac{200 m v}{t}

B

\dfrac{100 m v}{t}

C

\dfrac{m v}{100 t}

D

200 m v t

Correct Answer

Option A

Solution

When the balls strike the wall, the change in momentum of each ball is given by:

\Delta p = mv - (-mv) = 2mv

Since there are 100 balls, the total change in momentum of all the balls is

\Delta P = 2m(100v) = 200mv.

The time taken for all the balls to strike the wall is $\mathrm{t}$ seconds.

Therefore, the average force exerted on the wall is given by:

F = \frac{\Delta P}{\mathrm{t}} = \frac{2m(100v)}{\mathrm{t}} = \frac{200mv}{\mathrm{t}}

Therefore, the total force exerted by the balls on the wall is

\boxed{F = \frac{200mv}{\mathrm{t}}}

Q88

If momentum of a body is increased by 20%, then its kinetic energy increases by

A 36%

B 40%

C 44%

D 48%

Correct Answer

Option C

Solution

Let, initial momentum of body

({p_i}) = p

$\therefore$ Final momentum

({p_f}) = {p_i} + 20\%

of

{p_i}

= p + 0.2~p

= 1.2~p

We know, Kinetic energy

(E) = {{{p^2}} \over {2m}}

$\therefore$

{E_i} = {{{p^2}} \over {2m}}

and

{E_f} = {{{{(1.2p)}^2}} \over {2m}} = {{1.44\,{p^2}} \over {2m}}

$\therefore$ % Change in kinetic energy

= {{{E_f} - {E_i}} \over {{E_i}}} \times 100

= {{{{1.44\,{p^2}} \over {2m}} - {{{p^2}} \over {2m}}} \over {{{{p^2}} \over {2m}}}} \times 100

= {{{{{p^2}} \over {2m}}(1.44 - 1)} \over {{{{p^2}} \over {2m}}}} \times 100 = 0.44 \times 100 = 44

Q89

A rod of length L has non-uniform linear mass density given by

\rho

(x) =

a + b{\left( {{x \over L}} \right)^2}

, where a and b are constants and 0

\le

x

\le

L. The value of x for the centre of mass of the rod is at :

A

{3 \over 2}\left( {{{a + b} \over {2a + b}}} \right)L

B

{4 \over 3}\left( {{{a + b} \over {2a + 3b}}} \right)L

C

{3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L

D

{3 \over 2}\left( {{{2a + b} \over {3a + b}}} \right)L

Correct Answer

Option C

Solution

$\rho$ = a + b

{\left( {{x \over L}} \right)^2}

dm = $\rho$ dx =

\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx

M =

\int {dm}

=

\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx}

Xcom =

{{\int {xdm} } \over {\int {dm} }}

=

{{\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)xdx} } \over {\int\limits_0^L {\left( {a + b{{{x^2}} \over {{L^2}}}} \right)dx} }}

=

{{{{a{L^2}} \over 2} + {b \over {{L^2}}}.{{{L^4}} \over 4}} \over {aL + {b \over {{L^2}}}.{{{L^3}} \over 3}}}

=

{{\left( {{{4a + 2a} \over 8}} \right)L} \over {\left( {{{3a + b} \over 3}} \right)}}

=

{3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L

Q90

In two different experiments, an object of mass

5 \mathrm{~kg}

moving with a speed of

25 \mathrm{~ms}^{-1}

hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :

A Impulse and average force acting on the object will be same for both the cases.

B Impulse will be same for both the cases but the average force will be different.

C Average force will be same for both the cases but the impulse will be different.

D Average force and impulse will be different for both the cases.

Correct Answer

Option B

Solution

\Delta

P = impulse = same since acceleration is different force acting will be different.