A bullet with a mass of m1 = 10g with a speed of v = 300 m / s flew out of the barrel of the pistol.

A bullet with a mass of m1 = 10g with a speed of v = 300 m / s flew out of the barrel of the pistol. the bolt of a pistol with a mass of m2 = 200 g is pressed against the barrel by a spring, the stiffness of which is k = 25 kN / m. how far will the bolt move away after firing? consider that the pistol is rigidly fixed.

mp = 10 g = 0.01 kg.

Vp = 300 m / s.

mz = 200 g = 0.2 kg.

k = 25 kN / m = 25000 N / m.

x -?

Let’s find the shutter speed Vz immediately after the bullet has been fired from the law of conservation of momentum: mz * Vz = mp * Vp.

Vz = mp * Vp / mz.

According to the law of conservation of energy, the kinetic energy of the gate turns into potential energy: mz * Vz2 / 2 = k * x2 / 2.

x = √ (mz * Vz2) / √k = Vz * √mz / √k = mp * Vp * √mz / mz * √k.

x = 0.01 kg * 300 m / s * √0.2 kg / 0.2 kg * √25000 N / m = 0.04 m.

Answer: the bolt spring will be compressed by x = 0.04 m after firing.