QJEEM23Phae2b: JEE Main 2023 Physics - Class 11 JEEM23Phae2b Solution: A bullet of leaves the barrel of gun with a velocity of . If…

First, we need to find the velocity of the gun after the bullet is fired. We can use conservation of momentum to do this. The total momentum of the system of the gun and bullet is conserved before and after the bullet is fired. Therefore, we can write

$m_g u_g + m_b u_b = m_g v_g + m_b v_b$

where $u_g = 0$ is the initial velocity of the gun, $u_b = 600 \mathrm{~m/s}$ is the initial velocity of the bullet, $m_g = 3 \mathrm{~kg}$ is the mass of the gun, $m_b = 0.01 \mathrm{~kg}$ is the mass of the bullet, $v_b = 0$ is the final velocity of the bullet (since it has left the gun), and $v_g$ is the final velocity of the gun.

Substituting the given values, we get

$(3 \mathrm{~kg})(0) + (0.01 \mathrm{~kg})(600 \mathrm{~m/s}) = (3 \mathrm{~kg})v_g + (0.01 \mathrm{~kg})(0)$

Solving for $v_g$ , we get

$v_g = \frac{0.01 \mathrm{~kg} \times 600 \mathrm{~m/s}}{3 \mathrm{~kg}} = 2 \mathrm{~m/s}$

Therefore, the velocity of the gun after the bullet is fired is $v_g = 2 \mathrm{~m/s}$ .

Next, we need to find the impulse on the gun. The impulse-momentum theorem states that the impulse on an object is equal to the change in momentum of that object. Therefore, the impulse on the gun is given by

$I = \Delta p = m_g \Delta v$

where $\Delta v = v_f - u_g$ is the change in velocity of the gun. Since the initial velocity of the gun is zero, we can write $\Delta v = v_g$ .

Substituting the given values, we get

$I = (3 \mathrm{~kg}) \times (2 \mathrm{~m/s}) = 6 \mathrm{~Ns}$

Therefore, the impulse on the gun is $6 \mathrm{~Ns}$ , which is the correct answer.