Center of Mass and Collision — Page 5 | JEE Physics

Q41

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be

A 1.5 cm

B 2.0 cm

C 0.5 cm

D 1.0 cm

Correct Answer

Option D

Solution

Let's solve this step by step.

The original disc has a radius of 20 cm and is uniformly dense.

Its mass (or weight) is proportional to its area:

M = \pi \times 20^2 = 400\pi.

A hole of radius 5 cm is cut out. Similarly, its mass (if it were a complete disc) would be:

m = \pi \times 5^2 = 25\pi.

Since the hole cuts through the disc, we treat it as having a negative mass.

The center of the original disc is at the origin (0,0), and we need to locate the center of the hole.

The problem states that the edge of the hole touches the edge of the disc.

This means that the distance between the centre of the hole and the centre of the large disc is:

20 \text{ cm (disc radius)} - 5 \text{ cm (hole radius)} = 15 \text{ cm}.

For convenience, we can assume that the hole is positioned along the positive x-axis.

Thus, the centre of the hole is at:

\vec{r}_\text{hole} = (15, 0).

Now, the centre of mass (COM) of the remaining shape (the disc with the hole) can be calculated using the idea of superposition:

\vec{R}_{\text{cm}} = \frac{M \cdot \vec{r}_\text{disc} - m \cdot \vec{r}_\text{hole}}{M - m}.

Here, $\vec{r}_\text{disc} = (0,0)$ (since the large disc is centered at the origin). Plug in the values:

\vec{R}_{\text{cm}} = \frac{400\pi \cdot (0,0) - 25\pi \cdot (15,0)}{400\pi - 25\pi} = \frac{-(25\pi)(15,0)}{375\pi} = \frac{-375\pi \cdot (1,0)}{375\pi} = (-1,0).

This tells us that the centre of mass of the remaining piece is 1 cm from the origin, along the negative x-axis.

Thus, the distance of the centre of mass from the origin is:

\boxed{1.0\text{ cm}}.

So, the correct answer is Option D.

Q42

A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity

A

v

B

\dfrac{v}{3}

C

\dfrac{v}{4}

D

\dfrac{v}{2}

Correct Answer

Option B

Solution

When two particles collide, the total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum.

Therefore, we can write:

mv = (m+2m) v_f

where $v_f$ is the final velocity of the combined particles. Simplifying the above equation, we get:

v_f = \frac{mv}{3m} = \frac{v}{3}

So, the correct option is

\frac{v}{3}

.

Q43

An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is

A 2m

B 4m

C 1.5m

D 3.5m

Correct Answer

Option B

Solution

We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$ Let $\alpha$ particle rebounds with velocity $v_1$ , then Given : final energy of $\alpha=36 \%$ of initial energy

\begin{aligned} \Rightarrow \frac{1}{2} m v_1^2 =0.36 \times \frac{1}{2} m v^2 \\\\ \Rightarrow v_1 = 0.6 v .......(i) \end{aligned}

As unknown nucleus gained $64 \%$ of energy of $\alpha$ , we have

\begin{aligned} & \frac{1}{2} M v_2^2=0.64 \times \frac{1}{2} m v^2 \\ & \Rightarrow v_2=\sqrt{\frac{m}{M}} \times 0.8 v .........(ii) \end{aligned}

From momentum conservation, we have

m v=M v_2-m v_1

Substituting values of $v_1$ and $v_2$ from Eqs. (i) and (ii), we have

\begin{array}{rlrl} m v =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\\\ \Rightarrow 16 m v =\sqrt{m M} \times 0.8 v \\\\ \Rightarrow 2 m =\sqrt{m M} \\\\ \Rightarrow 4 m^2 =m M \Rightarrow M=4 m \end{array}

Q44

Two bodies of mass

4 \mathrm{~g}

and

25 \mathrm{~g}

are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is :

A

3: 5

B

5: 4

C

2: 5

D

4: 5

Correct Answer

Option C

Solution

\begin{aligned} & \frac{\mathrm{P}_1^2}{2 \mathrm{~m}_1}=\frac{\mathrm{P}_2^2}{2 \mathrm{~m}_2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\sqrt{\frac{\mathrm{m}_1}{\mathrm{~m}_2}}=\frac{2}{5} \end{aligned}

Q45

A body of mass

8 \mathrm{~kg}

and another of mass

2 \mathrm{~kg}

are moving with equal kinetic energy. The ratio of their respective momentum will be :

A 1 : 1

B 2 : 1

C 1 : 4

D 4 : 1

Correct Answer

Option B

Solution

P = \sqrt {2m\,KE}

\Rightarrow {{{P_1}} \over {{P_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}}

= \sqrt {{8 \over 2}} = {2 \over 1}

Q46

The mass of a hydrogen molecule is 3.32

\times

10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly:

A 2.35

\times

103 N m-2

B 4.70

\times

103 N m-2

C 2.35

\times

102 N m-2

D 4.70

\times

102 N m-2

Correct Answer

Option A

Solution

Considering one hydrogen molecule : As collision is elastic so, e = 1 Initial momentum,

\overrightarrow {{P_i}}

=

{{mv} \over {\sqrt 2 }}

\widehat i

$-$

{{mv} \over {\sqrt 2 }}

\widehat j

Final momentum,

\overrightarrow {{P_f}}

=

{{mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)

$-$

{{mv} \over {\sqrt 2 }}\widehat j

\therefore\,\,\,

Change in momentum for single H molecule,

\Delta

P =

\overrightarrow {{P_f}}

$-$

\overrightarrow {{P_i}}

=

{{2mv} \over {\sqrt 2 }}\left( { - \widehat i} \right)

\therefore\,\,\,

\left| {\Delta P} \right|

=

{{2mv} \over {\sqrt 2 }}

Now for n hydrogen molecule total momentum changes per second, =

\left( {{{2mv} \over {\sqrt 2 }}} \right)

$\times$ n As we know, Force (F) =

{{\Delta P} \over {\Delta t}}

=

{{{2mv} \over {\sqrt 2 }}}

$\times$ n

\therefore\,\,\,

As direction of

\Delta

P is towards negative i so force on the molecule will also be towards negative i direction.

From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.

\therefore\,\,\,

Force on the wall =

{{{2mv} \over {\sqrt 2 }}}

$\times$ n

\therefore\,\,\,

Pressure on the wall, P =

{F \over A}

=

{{2mv\,n} \over {\sqrt 2 A}}

=

{{2 \times 3.32 \times {{10}^{ - 27}} \times {{10}^3} \times {{10}^{23}}} \over {\sqrt 2 \times2\times {{10}^{ - 4}}}}

= 2.35 $\times$ 103 N m $-$ 2

Q47

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

Q48

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}

Q49

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}

Q50

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