A bullet of mass 10 g is fired horizontally with a velocity of 40 m/s from a gun of mass 8 kg. Using conservation of momentum, what is the recoil velocity (in m/s) of the gun?

Introduction / Context:
This question is a classic example of the law of conservation of linear momentum applied to a gun firing a bullet. When a bullet is fired forward, the gun recoils backward so that the total momentum of the system gun plus bullet remains constant, provided no external horizontal force acts. Such problems are common in basic mechanics and help students understand how momentum is shared between interacting bodies.

Given Data / Assumptions:

• Mass of bullet, m_bullet = 10 g = 0.01 kg.
• Velocity of bullet after firing, v_bullet = 40 m/s (forward direction).
• Mass of gun, m_gun = 8 kg.
• Initial velocity of both gun and bullet is zero; system is at rest before firing.
• No external horizontal force acts on the gun bullet system during the short firing interval.

Concept / Approach:
The law of conservation of linear momentum states that if no external force acts on a system, the total momentum of the system remains constant. Initially, the gun and bullet are at rest, so initial momentum is zero. After firing, the bullet moves forward and the gun recoils backward. The forward momentum of the bullet must be exactly balanced by the backward momentum of the gun so that the vector sum remains zero. We choose the forward direction as positive. If v_gun is the recoil velocity of the gun, then m_bullet * v_bullet + m_gun * v_gun = 0. Solving this relation gives the recoil velocity. The sign will indicate direction; the magnitude is asked in the question.

Step-by-Step Solution:
Step 1: Write the conservation of momentum equation: initial momentum = final momentum.Step 2: Initially, both gun and bullet are at rest, so initial momentum = 0.Step 3: After firing, bullet momentum = m_bullet * v_bullet = 0.01 * 40 = 0.4 kg m/s (forward).Step 4: Let the recoil velocity of the gun be v_gun (negative for backward motion). Then gun momentum = m_gun * v_gun = 8 * v_gun.Step 5: Apply conservation: 0 = total final momentum = 0.4 + 8 * v_gun.Step 6: Rearrange to get 8 * v_gun = -0.4, so v_gun = -0.4 / 8 = -0.05 m/s.Step 7: The negative sign shows backward direction; the magnitude of recoil velocity is 0.05 m/s.

Verification / Alternative check:
We can check the reasonableness of the result by comparing masses. The gun is much heavier than the bullet, so its recoil speed should be much smaller than the bullet speed. The bullet moves at 40 m/s, and the gun recoils at only 0.05 m/s, which is 800 times smaller. This ratio is consistent with the mass ratio 8 kg versus 0.01 kg, because momentum m * v must be balanced on both sides. Also, if we used another option, such as 2 m/s or 4 m/s, the recoil momentum of the gun would be huge compared to the bullet momentum, violating conservation. Therefore 0.05 m/s is the only value that keeps total momentum equal to zero.

Why Other Options Are Wrong:
For a recoil velocity of 2 m/s, gun momentum would be 8 * 2 = 16 kg m/s, far larger than the bullet momentum of 0.4 kg m/s, so total momentum would not be zero. For 0.1 m/s, gun momentum would be 8 * 0.1 = 0.8 kg m/s, still larger than the bullet momentum in magnitude. For 4 m/s, gun momentum becomes 32 kg m/s, which is clearly impossible in this context. Only 0.05 m/s gives gun momentum of 0.4 kg m/s backward, exactly balancing the bullet momentum forward.

Common Pitfalls:
Students sometimes forget to convert grams into kilograms and obtain wrong numerical values. Others may mix up the sign and think the recoil is in the same direction as the bullet, which would not conserve momentum. Another frequent mistake is to equate the velocities directly instead of momenta, ignoring the different masses. To avoid these errors, always convert units to SI, write the conservation equation with masses and velocities, and solve carefully.

Final Answer:
The recoil velocity of the gun has magnitude 0.05 m/s in the backward direction.