Center of Mass and Collision Questions — JEE Physics

Q1

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

Q2

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}

Q3

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

Q4

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

Q5

Particle A of mass m1 moving with velocity

\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}

collides with another particle B of mass m2 which is at rest initially. Let

\overrightarrow {{V_1}}

and

\overrightarrow {{V_2}}

be the velocities of particles A and B after collision respectively. If m1 = 2m2 and after collision

\overrightarrow {{V_1}} =

\left( {\widehat i + \sqrt 3 \widehat j} \right)

, the angle between

\overrightarrow {{V_1}}

and

\overrightarrow {{V_2}}

is :

A 105o

B 15o

C -45o

D 60o

Correct Answer

Option A

Solution

Given m1 = 2m2 So let, m2 = m and m1 = 2m From momentum conservation

{\overrightarrow p _i}

=

{\overrightarrow p _f}

$\Rightarrow$ (2m)

\left( {\sqrt 3 \widehat i + \widehat j} \right)

+ 0 = 2m

\left( {\widehat i + \sqrt 3 \widehat j} \right)

+ m

{\overrightarrow V _2}

$\Rightarrow$

{\overrightarrow V _2}

= 2

\left( {\sqrt 3 \widehat i + \widehat j} \right)

- 2

\left( {\widehat i + \sqrt 3 \widehat j} \right)

$\Rightarrow$

{\overrightarrow V _2}

=

\left( {2\sqrt 3 - 2} \right)\widehat i - \widehat j\left( {2\sqrt 3 - 2} \right)

=

2\left( {\sqrt 3 - 1} \right)\left( {\widehat i - \widehat j} \right)

Also given after collision

\overrightarrow {{V_1}} = \left( {\widehat i + \sqrt3\widehat j} \right)m{s^{ - 1}}

For angle between

{\overrightarrow V _1}

&

{\overrightarrow V _2}

, cos $\theta$ =

{{{{\overrightarrow V }_1}.{{\overrightarrow V }_2}} \over {\left| {{{\overrightarrow V }_1}} \right|\left| {{{\overrightarrow V }_2}} \right|}}

=

{{2\left( {\sqrt 3 - 1} \right) \times 1 - 2\left( {\sqrt 3 - 1} \right) \times \sqrt 3 } \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

=

{{2\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)} \over {2 \times 2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}

=

{{\left( {1 - \sqrt 3 } \right)} \over {2\sqrt 2 }}

$\Rightarrow$ $\theta$ = 105o

Q6

If 1022 gas molecules each of mass 10–26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 104 m/s, the pressure exerted by the gas molecules will be of the order of :

A 108 N/m2

B 1016 N/m2

C 104 N/m2

D 2 N/m2

Correct Answer

Option D

Solution

Momentum imparted to the surface in one collision,

\Delta p = ({p_i} - {p_f}) = mv - ( - mv) = 2mv

..... (i) Force on the surface due to n collision per second,

F = {n \over t}(\Delta p) = n\Delta p

( $\because$

t = 1s

) = 2 mnv [from Eq. (i)] So, pressure on the surface,

p = {F \over A} = {{2mnv} \over A}

Here, m = 10 $-$ 26 kg, n = 1022 s $-$ 1, v = 104 ms $-$ 1, A = 1 m2 $\therefore$ Pressure,

p = {{2 \times {{10}^{ - 26}} \times {{10}^{22}} \times {{10}^4}} \over 1} = 2

N/m2 So, pressure exerted is of order of 100.

Q7

A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the man with respect to the surface is :

A 0.28 ms–1

B 0.47 ms–1

C 0.20 ms–1

D 0.14 ms–1

Correct Answer

Option C

Solution

50 V1 = 20 V2 V1 + V2 = 0.70 V1 = 0.20

Q8

Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :

A 4 cm towards the 10 kg block

B 2 cm away from the 10 kg block

C 2 cm towards the 10 kg block

D 4 cm away from the 10 kg block

Correct Answer

Option C

Solution

For COM to remain unchanged, m1x1 = m2x2 $\Rightarrow$ 10 $\times$ 6 = 30 $\times$ x2 $\Rightarrow$ x2 = 2 cm towards 10 kg block.

Q9

If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

A 100%

B 200%

C 300%

D 400%

Correct Answer

Option A

Solution

K2 = 4K1

{1 \over 2}

mv

_2^2

= 4

{1 \over 2}

mv

_1^2

v2 = 2v1 P = mv P2 = mv2 = 2mv1 P1 = mv1 % change =

{{\Delta P} \over {{P_1}}} \times 100 = {{2m{v_1} - m{v_1}} \over {m{v_1}}} \times 100 = 100\%

Q10

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms

-

1 collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?

A 100 N

B 200 N

C 300 N

D 400 N

Correct Answer

Option B

Solution

Change in momentum of one ball = 2 $\times$ (0.05)(10) kg m/s = 1 kg m/s

\Rightarrow {F_{avg}} = {1 \over {\Delta t}} = {1 \over {0.005}}

N = 200 N