Center of Mass and Collision — Page 7 | NEET Physics

Q61

A ball of mass

200 \mathrm{~g}

rests on a vertical post of height

20 \mathrm{~m}

. A bullet of mass

10 \mathrm{~g}

, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance

30 \mathrm{~m}

and the bullet at a distance of

120 \mathrm{~m}

from the foot of the post. The value of initial velocity of the bullet will be (if

g=10 \mathrm{~m} / \mathrm{s}^{2}

) :

A 120 m/s

B 360 m/s

C 400 m/s

D 60 m/s

Correct Answer

Option B

Solution

$\because$ Time of flight of each ball and bullet

= \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2

s $\Rightarrow$ By applying linear momentum conservation

100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)

u = 360

m/s

Q62

A player caught a cricket ball of mass

150

g

moving at a rate of

20

m/s.

If the catching process is completed in

0.1s,

the force of the blow exerted by the ball on the hand of the player is equal to

A

150

N

B

3

N

C

30

N

D

300

N

Correct Answer

Option C

Solution

We know, Force $\times$ time = Impulse = Change in momentum $\therefore$

F \times t = m\left( {v - u} \right)

$\Rightarrow$

F = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N

Q63

A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :

A

2\sqrt {{k \over p}}

B

2\sqrt {{p \over k}}

C

\sqrt {{{2p} \over 2}}

D

\sqrt {{{2k} \over p}}

Correct Answer

Option B

Solution

{{dp} \over {dt}} = F = kt

\int_P^{3P} {dP} = \int_0^T {kt\,dt}

2p = {{K{T^2}} \over 2}

T = 2\sqrt {{P \over K}}

Q64

Two bodies of mass

1 \mathrm{~kg}

and

3 \mathrm{~kg}

have position vectors

\hat{i}+2 \hat{j}+\hat{k}

and

-3 \hat{i}-2 \hat{j}+\hat{k}

respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :

A

\hat{i}+2 \hat{j}+\hat{k}

B

-3 \hat{i}-2 \hat{j}+\hat{k}

C

-2 \hat{i}+2 \hat{k}

D

2 \hat{i}-\hat{j}+2 \hat{k}

Correct Answer

Option A

Solution

{\overline r _{com}} = {{{m_1}{{\overline r }_1} + {m_2}{{\overline r }_2}} \over {{m_1} + {m_2}}}

= {{(1 - 9)\widehat i + (2 - 6)\widehat j + (1 + 3)\widehat k} \over 4}

= {{ - 8\widehat i - 4\widehat j + 4\widehat k} \over 4}

{\overline r _{com}} = - 2\widehat i - \widehat j + \widehat k

\left| {\overline r } \right| = \sqrt {4 + 1 + 1} = \sqrt 6

\left| {\widehat i + 2\widehat j + \widehat k} \right| = \sqrt 6

Q65

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}

Q66

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 }}

Q67

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

Q68

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

Q69

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.

Q70

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}