Impulse–momentum (recoil) — a 0.2 kg bullet is fired horizontally at 25 m/s from a gun mounted on a carriage; the gun plus carriage mass is 100 kg. Find the recoil velocity of the gun–carriage system.

Introduction / Context:Gun recoil problems illustrate conservation of linear momentum. Before firing, the system is at rest; after firing, the forward momentum of the projectile is balanced by an equal and opposite momentum of the gun–carriage assembly (neglecting external impulses and resistances over the short firing interval).

Given Data / Assumptions:

• Bullet mass mb = 0.2 kg; bullet speed vb = 25 m/s (horizontal).
• Gun plus carriage mass Mg = 100 kg; recoil speed vg = ?
• System initially at rest; external horizontal impulse ≈ 0 over firing instant.

Concept / Approach:

Apply conservation of linear momentum along the line of fire: total momentum before equals total momentum after. The bullet acquires forward momentum; the gun recoils with equal magnitude opposite in direction.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Unit check: kg*m/s divided by kg gives m/s; magnitude is small due to large gun mass, as expected.

Why Other Options Are Wrong:

1.00 m/s and 1.5 m/s would require much lighter gun mass. 0.01 m/s or 0.005 m/s underestimate momentum balance.

Common Pitfalls (misconceptions, mistakes):

Using 200 g as 200 kg by unit error; adding rather than balancing momenta; ignoring sign convention (recoil direction is opposite).

Final Answer:

0.05 m/s