Answer to Question #110686 in Mechanics | Relativity for Rethabile Force

Question

A 0.010-kg bullet is fired from a 5.0-kg gun with a muzzle velocity of 230 m/s. Call the direction in which the bullet is fired from the +x direction .

Part A:
While the bullet is traveling in the barrel,what is the ratio of the acceleration of the gun to the acceleration of the bullet ?

Part B:
What is the speed of the bullet relative to the ground at the instant the bullet leaves the barrel?

Answer 1

$m_1v_1=m_2v_2\\0.01*230=5*v_2$

$v_2=0.46 m/s$

As the momentum on both sides is same and lets assume that the momentum change is in $\Delta t$ time

then force is equal to $\frac{\Delta P}{\Delta t }=ma$

$m_1a_1=m_2a_2$

$\frac{m_1}{m_2}=\frac{a_2}{a_1}$

$\frac{a_2}{a_1}=\frac{5}{0.01}=500$