Center of Mass and Collision — Page 6 | JEE Physics

Q51

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.

Q52

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.

Q53

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}}}

Q54

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

Q55

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

Q56

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.

Q57

A body

A

of mass

M

while falling vertically downloads under gravity breaks into two-parts; a body

B

of mass

{1 \over 3}

M

and a body

C

of mass

{2 \over 3}

M.

The center of mass of bodies

B

and

C

taken together shifts compared to that of bodies

B

and

C

taken together shifts compared to that of body

A

towards

A does not shift

B depends on height of breaking

C body

B

D body

C

Correct Answer

Option A

Solution

The center of mass does not shift as no external force is applied horizontally.

So the center of mass of the system continues its original path.

It is only the internal forces which comes into play while breaking.

Q58

An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2 : m1 is :

A 1 : 1

B 3 : 1

C 2 : 1

D 1 : 2

Correct Answer

Option B

Solution

Before collision After collision, Applying momentum conservation, m1v1 = m2v $-$ m1v .....

(1) As collision is elastic, $\therefore$ e = 1 $\Rightarrow$

{{v - ( - v)} \over {0 - {v_1}}} = 1

$\Rightarrow$ v1 = 2v .....

(2) Put value of v1 in equation (1), m1 (2v) = (m2 $-$ m1)v $\Rightarrow$ 2m1 = m2 $-$ m1 $\Rightarrow$ 3m1 = m2 $\Rightarrow$

{{{m_2}} \over {{m_1}}} = {3 \over 1}

Q59

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

A Momentum

B Potential Energy

C Acceleration

D Force

Correct Answer

Option A

Solution

At maximum height it's velocity is zero. So momentum (mv) will be zero.

Q60

A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s2 . Assume there is no rotational motion and loss of energy after the collision is negligable.]

A 23 J

B 21 J

C 20 J

D 19 J

Correct Answer

Option B

Solution

Let velocity of block after collision = v Using momentum conservation, pi = pf 0.1×20 = (1.9 + 0.1)V $\Rightarrow$ 2 = 2 V $\Rightarrow$ V = 1 m/sec Kinetic energy just before striking the floor =

{1 \over 2}m{v^2} + mgh

=

{1 \over 2} \times 2{\left( 1 \right)^2} + 2 \times 10 \times 1

= 21 J