Center of Mass and Collision — Page 10 | NEET Physics

Q91

A body

A

of mass

M

while falling vertically downloads under gravity breaks into two-parts; a body

B

of mass

{1 \over 3}

M

and a body

C

of mass

{2 \over 3}

M.

The center of mass of bodies

B

and

C

taken together shifts compared to that of bodies

B

and

C

taken together shifts compared to that of body

A

towards

A does not shift

B depends on height of breaking

C body

B

D body

C

Correct Answer

Option A

Solution

The center of mass does not shift as no external force is applied horizontally.

So the center of mass of the system continues its original path.

It is only the internal forces which comes into play while breaking.

Q92

An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2 : m1 is :

A 1 : 1

B 3 : 1

C 2 : 1

D 1 : 2

Correct Answer

Option B

Solution

Before collision After collision, Applying momentum conservation, m1v1 = m2v $-$ m1v .....

(1) As collision is elastic, $\therefore$ e = 1 $\Rightarrow$

{{v - ( - v)} \over {0 - {v_1}}} = 1

$\Rightarrow$ v1 = 2v .....

(2) Put value of v1 in equation (1), m1 (2v) = (m2 $-$ m1)v $\Rightarrow$ 2m1 = m2 $-$ m1 $\Rightarrow$ 3m1 = m2 $\Rightarrow$

{{{m_2}} \over {{m_1}}} = {3 \over 1}

Q93

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

A Momentum

B Potential Energy

C Acceleration

D Force

Correct Answer

Option A

Solution

At maximum height it's velocity is zero. So momentum (mv) will be zero.

Q94

A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s2 . Assume there is no rotational motion and loss of energy after the collision is negligable.]

A 23 J

B 21 J

C 20 J

D 19 J

Correct Answer

Option B

Solution

Let velocity of block after collision = v Using momentum conservation, pi = pf 0.1×20 = (1.9 + 0.1)V $\Rightarrow$ 2 = 2 V $\Rightarrow$ V = 1 m/sec Kinetic energy just before striking the floor =

{1 \over 2}m{v^2} + mgh

=

{1 \over 2} \times 2{\left( 1 \right)^2} + 2 \times 10 \times 1

= 21 J

Q95

A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms

-

1, the impulse imparted to the gun and velocity of recoil of gun are :

A 0.2 kg ms

-

1, 0.1 ms

-

1

B 0.4 kg ms

-

1, 0.05 ms

-

1

C 0.2 kg ms

-

1, 0.05 ms

-

1

D 0.4 kg ms

-

1, 0.1 ms

-

1

Correct Answer

Option C

Solution

mBullet = 4g, MGun = 4 kg vBullet

\simeq

50 m/s Now, PB = Pg Pg = m $\times$ vBullet =

{4 \over {1000}}

$\times$ 50 = 0.2 kg m/s So impulse = 0.2 kg m/s

{v_G} = {{0.2} \over {{M_{Gun}}}} = {{0.2} \over 4} = 0.05

m/s

Q96

An average force of

125 \mathrm{~N}

is applied on a machine gun firing bullets each of mass

10 \mathrm{~g}

at the speed of

250 \mathrm{~m} / \mathrm{s}

to keep it in position. The number of bullets fired per second by the machine gun is :

A 25

B 50

C 5

D 100

Correct Answer

Option B

Solution

To find the number of bullets fired per second, we can use the concept of momentum.

When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum.

The force applied on the machine gun is used to balance this backward force.

First, let's find the momentum of each bullet: Momentum = Mass × Velocity Bullet mass =

10 \mathrm{~g} = 0.01 \mathrm{~kg}

Bullet velocity =

250 \mathrm{~m/s}

Momentum per bullet =

0.01 \mathrm{~kg} \cdot 250 \mathrm{~m/s} = 2.5 \mathrm{~kg \cdot m/s}

Now let's find the momentum per second that needs to be balanced by the applied force: Force =

125 \mathrm{~N}

Momentum per second = Force × Time Since we are considering a time interval of 1 second, the momentum per second is equal to the applied force: Momentum per second =

125 \mathrm{~N}

Now, let's find the number of bullets fired per second: Number of bullets = Momentum per second / Momentum per bullet Number of bullets =

\frac{125 \mathrm{~N}}{2.5 \mathrm{~kg \cdot m/s}}

Number of bullets = 50 Therefore, the machine gun fires 50 bullets per second.

Q97

A spherical body of mass

100 \mathrm{~g}

is dropped from a height of

10 \mathrm{~m}

from the ground. After hitting the ground, the body rebounds to a height of

5 \mathrm{~m}

. The impulse of force imparted by the ground to the body is given by : (given,

\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2

)

A

43.2 \mathrm{~kg} \mathrm{~ms}^{-1}

B

2.39 \mathrm{~kg} \mathrm{~ms}^{-1}

C

4.32 \mathrm{~kg} \mathrm{~ms}^{-1}

D

23.9 \mathrm{~kg} \mathrm{~ms}^{-1}

Correct Answer

Option B

Solution

\begin{aligned} \vec{I} & =\Delta \vec{P}=\vec{P}_f-\vec{P}_i \\ \mathrm{M} & =0.1 \mathrm{~kg} \\ I & =\Delta P=0.1(\sqrt{2 \times 9.8 \times 5}-(-\sqrt{2 \times 9.8 \times 10})) \\ & =0.1(14+7 \sqrt{2}) \approx 2.39 \mathrm{~kg} \mathrm{~ms}^{-1} \end{aligned}

Q98

An artillery piece of mass

M_1

fires a shell of mass

M_2

horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

A

M_1 /\left(M_1+M_2\right)

B

\dfrac{M_2}{M_1}

C

\dfrac{M_1}{M_2}

D

M_2 /\left(M_1+M_2\right)

Correct Answer

Option B

Solution

\begin{aligned} & \left|\overrightarrow{\mathrm{p}_1}\right|=\left|\overrightarrow{\mathrm{p}_2}\right| \\ & \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{M}} ; \mathrm{p} \text{ same } \\ & \mathrm{KE} \propto \frac{1}{\mathrm{~m}} \\ & \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{\mathrm{p}^2 / 2 \mathrm{M}_1}{\mathrm{p}^2 / 2 \mathrm{M}_2}=\frac{\mathrm{M}_2}{\mathrm{M}_1} \end{aligned}