Center of Mass and Collision — Page 4 | JEE Physics

Q31

Two identical particles move towards each other with velocity

2v

and

v

respectively. The velocity of center of mass is

A

v

B

v/3

C

v/2

D zero

Correct Answer

Option C

Solution

The velocity of center of mass of two particle system is

{v_c} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}

= {{m\left( {2v} \right) + m\left( { - v} \right)} \over {m + m}}

= {v \over 2}

Q32

A particle of mass m with an initial velocity

u\widehat i

collides perfectly elastically with a mass 3 m at rest. It moves with a velocity

v\widehat j

after collision, then, v is given by :

A

v = \sqrt {{2 \over 3}} u

B

v = {u \over {\sqrt 3 }}

C

v = {u \over {\sqrt 2 }}

D

v = {1 \over {\sqrt 6 }}u

Correct Answer

Option C

Solution

From momentum conservation m(u)

\widehat i

+ 3m(0) = mv

\widehat j

+ 3m

\overrightarrow {v'}

$\Rightarrow$ 3m

\overrightarrow {v'}

= m(u)

\widehat i

- mv

\widehat j

$\Rightarrow$

\overrightarrow {v'} = {{u\widehat i - v\widehat j} \over 3}

$\therefore$

\left| {\overrightarrow {v'} } \right| = {{\sqrt {{u^2} + {v^2}} } \over 3}

$\Rightarrow$

{\left| {\overrightarrow {v'} } \right|^2} = {{{u^2} + {v^2}} \over 9}

......(1) As collision is perfectely elastic hence KEi = KEj $\Rightarrow$

{1 \over 2}m{u^2} + {1 \over 2}\left( {3m} \right) \times {0^2} = {1 \over 2}m{v^2} + {1 \over 2}3m{\left( {v'} \right)^2}

$\Rightarrow$ u2 = v2 + 3

{\left( {v'} \right)^2}

$\Rightarrow$ u2 = v2 + 3

\left( {{{{u^2} + {v^2}} \over 9}} \right)

$\Rightarrow$ 3u2 = 3v2 + u2 + v2 $\Rightarrow$ 2u2 = 4v2 $\Rightarrow$

v = {u \over {\sqrt 2 }}

Q33

A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25

\times

106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2 , the value of x will be close to :

A 8 cm

B 4 cm

C 40 cm

D 80 cm

Correct Answer

Option B

Solution

velocity of 1 kg block just before it collides with 3kg block =

\sqrt {2gh} = \sqrt {2000}

m/s Applying momentum conversation just before and just after collision. 1 $\times$

\sqrt {2000}

= 4v $\Rightarrow$ v =

{{\sqrt {2000} } \over 4}

m/s initial compression of spring 1.25 $\times$ 106 x0 = 30 $\Rightarrow$ x0 $\approx$ 0 applying work energy theorem, Wg + Wsp =

\Delta

KE $\Rightarrow$ 40 $\times$ x +

{1 \over 2}

$\times$ 1.25 $\times$ 106 (02 $-$ x2) = 0 $-$

{1 \over 2}

$\times$ 4 $\times$ v2 solving x $\approx$ 4 cm

Q34

A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - (g = 10 ms–2)

A 30 m

B 40 m

C 20 m

D 10 m

Correct Answer

Option B

Solution

Time taken for the particles to collide, t =

{f \over {{V_{rel}}}} = {{100} \over {100}} = 1

sec Speed of wood just before collision = gt = 10 m/s & speed of bullet just before collision v-gt = 100 $-$ 10 = 90 m/s Now, conservation of linear momentum just before and after the collision - $-$ (0.02) (1v) + (0.02) (9v) = (0.05)v $\Rightarrow$ 150 = 5v $\Rightarrow$ v = 30 m/s Max. height reached by body h =

{{{v^2}} \over {2g}}

h =

{{30 \times 30} \over {2 \times 10}}

= 45m $\therefore$ Height above tower = 40 m

Q35

Statement - 1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement - 2 : Principle of conservation of momentum holds true for all kinds of collisions.

A Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1

B Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1

C Statement - 1 is false, Statement - 2 is true

D Statement - 1 is true, Statement - 2 is false

Correct Answer

Option A

Solution

In completely inelastic collision,

{m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v

after collision both particle have common velocity

v

, so all energy is not lost $\therefore$ Statement -

1

is true The principle of conservation of momentum applicable for all kinds of collisions. So statement -

2

is also true. Statement -

2

explains statement -

1

correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.

Q36

A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by :-

A

{{{v^2}} \over {g}}

B

{{2{v^2}} \over {7g}}

C

{{{v^2}} \over {2g}}

D

{{2{v^2}} \over {5g}}

Correct Answer

Option D

Solution

Initial condition can be shown in the figure below As mass $m$ collides with wedge, let both wedge and mass move with speed $v^{\prime}$ .

Then, Given: Mass of the wedge (M) = 4m Mass of the particle (m) Initial speed of the particle (v) There is no friction.

Step 1 : Conservation of Linear Momentum Before the collision, the momentum of the system is just the momentum of the particle because the wedge is at rest.

After the collision, both the particle and the wedge will be moving.

Let's denote the final common velocity as $v'$ .

We can write the conservation of momentum as :

mv = (m + 4m)v'

which simplifies to

v' = \frac{v}{5} \tag{1}

Step 2 : Conservation of Mechanical Energy We apply the conservation of energy before and after the collision.

Before the collision, only the particle has kinetic energy.

After the collision, both the particle and the wedge have kinetic energy and the particle has potential energy due to its height h on the wedge.

We can write the conservation of energy as :

\frac{1}{2}mv^2 = \frac{1}{2}(m + 4m){v'}^2 + mgh

which simplifies to

mv^2 = (m + 4m){v'}^2 + 2mgh

Substituting equation (1) into this equation gives

v^2 = 5{v'}^2 + 2gh \Rightarrow \frac{4}{5}v^2 = 2gh

which simplifies to

h = \frac{2v^2}{5g}

So, the maximum height climbed by the particle on the wedge is given by $\dfrac{2v^2}{5g}$ .

Therefore, the correct answer is Option D :

\frac{2v^2}{5g}

Q37

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}

Q38

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

Q39

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}

Q40

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}}}