A gun of mass 3000 kg fires horizontally a shell of mass 50 kg with a velocity of 300 m/s. What is the velocity with which the gun will recoil?

Question

Question

This question was previously asked in

UPPSC AE Mechanical 2019 Official Paper I (Held on 13 Dec 2020)

Answer 1

Concept:

From momentum conservation we have,

initial momentum = final momentum i.e., p_i = p_f

0 = m_gv_g + m_sv_s [∵ initial velocity is zero]

m_gv_g = - m_sv_s

where,

m_g,v_g is the mass of the gun and velocity of gun respectively.

m_s,v_s is the mass of the shell and velocity of shell respectively

Calculation:

Given:

m_g= 3000 kg, m_s = 50 kg, v_s = 300 m/s

\({v_g} = - \frac{{{m_s}{v_s}}}{{{m_g}}}\)

\({v_g} = - \frac{{50 \times 300}}{{3000}} = - 5{\rm{ m/s}}\)