Question

from a rifle of mass 4 kg A bullet of mass 50 kg 5 horizontal with an initial velocity of 40 ms -¹calculate the initial recoil velocity of the rifle​

Answer 1

Answer:

We know that,

As per the principle of law of conservation of momentum,

m1u1 = m2v2

So, according to question,

M1 = 4 kg

M2 = 50 kg

V2 = 40 m/s

So, 4 * u = 50 * 40

4u = 2000

u = 2000/4 = 500 m/s

therefore initial recoil of gun will be 500 m/s as per the question given....

Explanation:

Answer 2

$\underline{ \underline{ \bf{Answer}}} : - \\ \implies \: \bf{recoil \: velocity \: = 0.05 \: \frac{m}{s} } \\ \\ \underline{\underline{\bf {Step - by - step \: explanation}}} : - \\ \\ \underline{\bf{ To \: find}} - \\ \\ \bf{find \: recoil \: velocity \: of \: rifle} \\ \\ \underline{\bf{Solution}} : - \\ \\ \bf{velocity \: of \: \: bullet \:( v1) = 40 \: \frac{m}{s} } \\ \\ \bf{ mass \: of \: bullet \: (m1) = 50 \: g = 0.050 \: kg} \\ \\ \bf{mass \: of \: rifle \: (m2) = 4 \: kg } \\ \\ \bf{recoil \: velocity \: of \: rifle \: (v2) =? }$

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

Momentum of rifle = (momentum of bullet)

$\implies \: \bf{m2 \: v2 = m1 \: v1} \\ \\ \implies \: \bf{4 \times v2 = 0.050 \times 40} \\ \\ \implies \: \bf{v2 = 0.44 \: \frac{m}{s} } \\ \\ \bf{recoil \: velocity \: of \: rifle \: is \: 0.05 \: \frac{m}{s} }$