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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A man fires a bullet of mass 200 g200 \text{ g} at a speed of 5 m/s5 \text{ m/s}. The gun is of one kg mass. By what velocity the gun rebounds backwards?

A

0.1 m/s0.1 \text{ m/s}

B

10 m/s10 \text{ m/s}

C

1 m/s1 \text{ m/s}

D

0.01 m/s0.01 \text{ m/s}

Step-by-Step Solution

According to the law of conservation of linear momentum, the total momentum of an isolated system remains constant. Before firing, both the gun and the bullet are at rest, so the initial momentum of the system is zero .

After firing, the final momentum must also be zero: Pinitial=PfinalP_{\text{initial}} = P_{\text{final}} 0=mbvb+mgvg0 = m_b v_b + m_g v_g

Given: Mass of bullet, mb=200 g=0.2 kgm_b = 200 \text{ g} = 0.2 \text{ kg} Velocity of bullet, vb=5 m/sv_b = 5 \text{ m/s} Mass of gun, mg=1 kgm_g = 1 \text{ kg} Recoil velocity of gun =vg= v_g

Substituting the values into the equation: 0=(0.2 kg×5 m/s)+(1 kg×vg)0 = (0.2 \text{ kg} \times 5 \text{ m/s}) + (1 \text{ kg} \times v_g) 0=1+vg0 = 1 + v_g vg=1 m/sv_g = -1 \text{ m/s}

The negative sign indicates that the gun rebounds in the opposite direction to the bullet. Therefore, the magnitude of the recoil velocity is 1 m/s1 \text{ m/s}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERbulletvelocityreboundsbackwards

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