Center of Mass and Collision — Page 7 | JEE Physics

Q61

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

Q62

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.

Q63

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}

Q64

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}