Question

From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s1. Calculate the initial recoil velocity of the rifle

Answer 1

Answer:

given m1=4000grms

m2=50grms

v1=35ms then the final velocity v2=

by law of conservation of mass

m1v1=m2v2

4000grms ×35=50×v2=v2=4000×35÷50=2800m/s^1

Answer 2

$\underline{ \underline{ \bf{Answer}}} : - \\ \implies \: \bf{recoil \: velocity \: = 0.44 \: \frac{m}{s} } \\ \\ \underline{\underline{\bf {Step - by - step \: explanation}}} : - \\ \\ \underline{\bf{ To \: find}} - \\ \\ \bf{find \: recoil \: velocity \: of \: rifle} \\ \\ \underline{\bf{Solution}} : - \\ \\ \bf{velocity \: of \: \: bullet \:( v1) = 35 \: \frac{m}{s} } \\ \\ \bf{ mass \: of \: bullet \: (m1) = 50 \: g = 0.050 \: kg} \\ \\ \bf{mass \: of \: rifle \: (m2) = 4 \: kg } \\ \\ \bf{recoil \: velocity \: of \: rifle \: (v2) =? }$

Since,No external force acts on the rifle then the change in momentum is zero.

Momentum of rifle = (momentum of bullet)

$\implies \: \bf{m2 \: v2 = m1 \: v1} \\ \\ \implies \: \bf{4 \times v2 = 0.050 \times 35} \\ \\ \implies \: \bf{v2 = 0.44 \: \frac{m}{s} } \\ \\ \bf{recoil \: velocity \: of \: rifle \: is \: 0.44 \: \frac{m}{s} }$