Center of Mass and Collision — Page 4 | NEET Physics

Q31

A 0.5 kg ball moving with a speed of 12 m/s strikes a hand wall at an angle of 30 o with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is :

A 96 N

B 48 N

C 24 N

D 12 N

Correct Answer

Option C

Solution

Resolving the velocities in vertical and horizontal directions, resolved parts of first velocity v cos $\theta$ perpendicular to the wall and v sinq parallel to the wall.

In the second case, they are –v sin $\theta$ & v cos $\theta$ respectively.

Here, –ve sign is because direction is opposite to the earlier ones.

So we see a net change in velocity perpendicular to way = v sin $\theta$ – (–v sin $\theta$ ) = 2v sin $\theta$ This change has occured in 0.25 sec, so, rate of change of velocity =

{{2v\sin \theta } \over {0.25}}

= {{2 \times 12 \times \sin 30^\circ } \over {0.25}} \Rightarrow {{24 \times 1} \over {2 \times 0.25}} = 48

Thus, acceleration a = 48 m/sec 2 Force applied = m $\times$ a = 0.5 × 48 = 24 N

Q32

A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 m s

-

1 . The kinetic energy of the other mass is :

A 324 J

B 486 J

C 256 J

D 524 J

Correct Answer

Option B

Solution

As per law of conservation of linear momentum m 1 v 1 + m 2 v 2 = 0

\Rightarrow {v_2} = \left( { - {{{m_1}} \over {{m_2}}}} \right){v_1}

\Rightarrow {v_2} = \left( { - {{18} \over {12}}} \right)6 = - 9\,m/s

Now, kinetic energy of second piece

K.E = {1 \over 2}\left( {mv_2^2} \right)

= {1 \over 2} \times 12 \times {\left( { - 9} \right)^2}

$\therefore$ K.E = 486 J

Q33

Consider a system of two particles having masses m 1 and m 2 . If the particle of mass m 1 is pushed towards the mass centre of particles through a distance d. by what distance would be particle of mass m 2 move so as to keep the mass centre of particles at the original position?

A

{{{m_1}} \over {{m_1} + m{}_2}}

d

B

{{{m_1}} \over {{m_2}}}

d

C d

D

{{{m_2}} \over {{m_1}}}

d

Correct Answer

Option B

Solution

C.M = {{{m_1}{x_1} + {m_2}{x_2}} \over {{m_1} + {m_2}}}

...(i) After changing position of m 1 and to keep the position of C.M. same

C.M = {{{m_1}\left( {{x_1} - d} \right) + {m_2}\left( {{x_2} + {d_2}} \right)} \over {{m_1} + {m_2}}}

\Rightarrow 0 = {{{m_1}d + {m_2}{d_2}} \over {{m_1} + {m_2}}}

[Substituting value of C.M. from (i)]

\Rightarrow {d_2} = {{{m_1}} \over {{m_2}}}d

Q34

A stationary particle explodes into two particles of masses m 1 and m 2 which move in opposite directions with velocities v 1 and v 2 . The ratio of their kinetic energies E 1 /E 2 is :

A m 2 /m 1

B m 1 /m 2

C 1

D m 1 v 2 /m 2 v 1

Correct Answer

Option A

Solution

From conservation law of momentum, before collision and after collision linear momentum (p) will be same.

That is, initial momentum = final momentum. $\Rightarrow$ 0 = m 1 v 1 – m 2 v 2 $\Rightarrow$ m 1 v 1 = m 2 v 2 p 1 = p 2 Now,

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

$\therefore$

{{{E_1}} \over {{E_2}}} = {{p_1^2} \over {2{m_1}}} \times {{2{m_2}} \over {p_2^2}}

\Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}\left[ {{p_1} = {p_2}} \right]

Q35

A rod of length is 3 m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at :

A 1.5 m

B 2 m

C 2.5 m

D 3.0 m.

Correct Answer

Option B

Solution

Let us consider an elementary length dx at a distance x from one end.

It’s mass = k ·x ·dx [k = proportionality constant] Then centre of gravity of the rod x c is given by

{x_c} = {{\int\limits_0^3 {kxdx.x} } \over {\int\limits_0^3 {kxdx} }} = {{\int\limits_0^3 {{x^2}dx} } \over {\int\limits_0^3 {xdx} }} = {{\left. {{{{x^3}} \over 3}} \right|_0^3} \over {\left. {{{{x^2}} \over 2}} \right|_0^3}}

\Rightarrow {x_c} = {{{{27} \over 3}} \over {{9 \over 2}}} = 2

$\therefore$ Centre of gravity of the rod will be at distance of 2 m from one end.

Q36

If kinetic energy of a body is increased by 300% then percentage change in momentum will be :

A 100%

B 150%

C 265%

D 73.2%

Correct Answer

Option A

Solution

Let m be the mass of the body and v 1 and v 2 be the initial and final velocities of the body respectively.

$\therefore$ Initial kinetic energy =

{1 \over 2}mv_1^2

Final kinetic energy =

{1 \over 2}mv_2^2

Initial kinetic energy is increased 300% to get the final kinetic energy. $\therefore$

{1 \over 2}mv_2^2 = {1 \over 2}\left( {1 + {{300} \over {100}}} \right)mv_1^2

$\Rightarrow$ v 2 = 2v 1 or v 2 /v 1 = 2 ... (i) Initial momentum = p 1 = mv 1 Final momentum = p 2 = mv 2 $\therefore$

{{{p_2}} \over {{p_1}}} = {{m{v_2}} \over {m{v_1}}} = {{{v_2}} \over {{v_1}}} = 2

$\therefore$

{p_2} = 2{p_1} = \left( {1 + {{100} \over {100}}} \right){p_1}

So momentum has increased 100%.

Q37

A bomb of mass

16kg

at rest explodes into two pieces of masses

4

kg

and

12

kg.

The velocity of the

12

kg

mass is

4\,\,m{s^{ - 1}}.

The kinetic energy of the other mass is

A

144

J

B

288

J

C

192

J

D

96

J

Correct Answer

Option B

Solution

Here linear momentum is conserved as no external force is acting on the bomb. Let the velocity and mass of

4

kg

piece be

{v_1}

and

{m_1}

and that of

12

kg

piece be

{v_2}

and

{m_2}

. Applying conservation of linear momentum

0 = {m_2}{v_2} - {m_1}{v_1}

\Rightarrow {v_1} = {{12 \times 14} \over 4} = 12\,m{s^{ - 1}}

$\therefore$

K.E{_1} = {1 \over 2}{m_1}v_1^2 = {1 \over 2} \times 4 \times 144 = 288\,J

Q38

Distance of the center of mass of a solid uniform cone from its vertex is

z{}_0

. If the radius of its base is

R

and its height is

h

then

z{}_0

is equal to :

A

{{5h} \over 8}

B

{{3{h^2}} \over {8R}}

C

{{{h^2}} \over {4R}}

D

{{3h} \over 4}

Correct Answer

Option D

Solution

Let the density of solid cone $\rho$ .

dm = \rho \pi {r^2}dy

{y_{cm}} = {{\int {ydm} } \over {\int {dm} }}

= {{\int\limits_0^h {\pi {r^2}} dy\rho \times y} \over {{1 \over 3}\pi {R^2}h\rho }}

= {{3h} \over 4}

Q39

A body of mass m1 moving with an unknown velocity of

{v_1}\mathop i\limits^ \wedge

, undergoes a collinear collision with a body of mass m2 moving with a velocity

{v_2}\mathop i\limits^ \wedge

. After collision, m1 and m2 move with velocities of

{v_3}\mathop i\limits^ \wedge

and

{v_4}\mathop i\limits^ \wedge

, respectively. If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :-

A

{v_4} - {{{v_2}} \over 2}

B

{v_4} - {{{v_2}} \over 4}

C

{v_4} - {v_2}

D

{v_4} + {v_2}

Correct Answer

Option C

Solution

Applying linear momentum conservation

{m_1}{v_1}\widehat i + {m_2}{v_2}\widehat i = {m_1}{v_3}\widehat i + {m_2}{v_4}\widehat i

m1v1 + 0.5 m1v2 = m1(0.5 v1) + 0.5 m1v4 0.5 m1v1 = 0.5 m1(v4 - v2) v1 = v4 - v2

Q40

A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be :-

A 2

\sqrt2

v

B v / (2

\sqrt2

)

C v /

\sqrt2

D

\sqrt2

v

Correct Answer

Option A

Solution

Initial momentum · Pi = 2mv + 2mv = 4 mv Let v' be the speed of

l

particle

\therefore 2{{mv} \over {\sqrt 2 }} = 4mv

$\Rightarrow$ v' =

{2\sqrt 2 }

v