Center of Mass and Collision — Page 8 | NEET Physics

Q71

A particle of mass m is projected with a speed u from the ground at an angle

\theta = {\pi \over 3}

w.r.t. horizontal (x-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity

u\widehat i

. The horizontal distance covered by the combined mass before reaching the ground is:

A

2\sqrt 2 {{{u^2}} \over g}

B

{{3\sqrt 3 } \over 8}{{{u^2}} \over g}

C

{{3\sqrt 2 } \over 4}{{{u^2}} \over g}

D

{5 \over 8}{{{u^2}} \over g}

Correct Answer

Option B

Solution

By momentum conservation,

{{\mu} \over 2}

+ mu = 2mV $\Rightarrow$ V =

{{3u} \over 4}

Hmax =

{{{u^2}{{\sin }^2}60^\circ } \over {2g}}

=

{{{u^2} \times {3 \over 4}} \over {2g}}

=

{{3{u^2}} \over {8g}}

Time taken =

\sqrt {{{2{H_{\max }}} \over g}}

=

\sqrt {{2 \over g} \times {{3{u^2}} \over {8g}}}

=

{{\sqrt 3 } \over 2}{u \over g}

Horizontal distance traveled = ut =

{{3u} \over 4}.{{\sqrt 3 } \over 2}{u \over g}

=

{{3\sqrt 3 } \over 8}{{{u^2}} \over g}

Q72

A machine gun fires a bullet of mass

40

g

with a velocity

1200m{s^{ - 1}}.

The man holding it can exert a maximum force of

144

N

on the gun. How many bullets can he fire per second at the most?

A Two

B Four

C One

D Three

Correct Answer

Option D

Solution

Assume the man can fire

n

bullets in one second. $\therefore$ change in momentum per second

= n \times mv = F

[

m=

mass of bullet,

v=

velocity,

F

= force) ] $\therefore$

n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3

Q73

Two bodies A and B of masses 5 kg and 8 kg are moving such that the momentum of body B is twice that of the body A. The ratio of their kinetic energies will be :

A 4 : 5

B 2 : 5

C 5 : 4

D 5 : 2

Correct Answer

Option B

Solution

Given, Mass of body A = 5 kg Mass of body B = 8 kg Momentum of body B is twice that of body A, $\therefore$

{P_B} = 2{P_A}

We know, Kinetic Energy

(K) = {{{P^2}} \over {2m}}

$\therefore$

{{{K_A}} \over {{K_B}}} = {\left( {{{{P_A}} \over {{P_B}}}} \right)^2} \times {{{m_B}} \over {{m_A}}}

= {\left( {{1 \over 2}} \right)^2} \times {8 \over 5}

= {1 \over 4} \times {8 \over 5}

= {2 \over 5}

Q74

A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of

\sqrt {2gh}

. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of

\sqrt {{h \over g}}

is :

A

\sqrt {{1 \over 2}}

B

{1 \over 2}

C

\sqrt {{3 \over 2}}

D

\sqrt {{3 \over 4}}

Correct Answer

Option C

Solution

Particles will collide after time t =

{h \over {\sqrt {2gh} }}

=

\sqrt {{h \over {2g}}}

Velocity of (A) just before collision, VA = 0 + gt =

g\sqrt {{h \over {2g}}}

=

\sqrt {{{gh} \over 2}}

Velocity of (B) just before collision, VB =

{\sqrt {2gh} }

-

g\sqrt {{h \over {2g}}}

=

{\sqrt {2gh} }

-

\sqrt {{{gh} \over 2}}

=

\sqrt {gh} \left[ {\sqrt 2 - {1 \over {\sqrt 2 }}} \right]

As 'mg' is non-impulsive so conserving linear momentum just before and just after collision, Pi = Pf $\Rightarrow$ m

\sqrt {gh} \left[ {\sqrt 2 - {1 \over {\sqrt 2 }}} \right]

- m

\sqrt {{{gh} \over 2}}

=

2m{V_f}

$\Rightarrow$ Vf = 0 height from ground where collision takes place = h1 = h -

{1 \over 2}g{t^2}

= h -

{1 \over 2}g.{h \over {2g}}

=

{{3h} \over 4}

Time taken by combined mass to reach the ground =

\sqrt {{{2 \times {{3h} \over 4}} \over {2g}}}

=

\sqrt {{{3h} \over {2g}}}

Q75

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.

Q76

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}

.

Q77

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}

Q78

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}

Q79

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}

Q80

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