Question

A bullet of mass 10 g is fired from a gun of mass 5.0 kg. If the velocity of bullet is 250m/s. Find the recoil velocity of the gun.

Answer 1

Details of the system before firing :-

m1 = 10g = 0.01 kg ; u1 = 0m/s
m2 = 5kg ; u2 = 0m/s

Hence,
Total momentum of the system before firing :-

$m1u1 + m2u2 \: = 0.01 \times 0 + 5 \times 0 = 0$
Now,
Details of the system after firing :-

m1 = 10g=0.01kg ; v1 = 250m/s
m2 =5kg ; v2 = ? ( v2 is the recoil velcoity of the gun)

Hence,
Total momentum of the system after firing,

$m1v1 + m2v2 = 0.01 \times 250 + 5 \times v2$
$m1v1 + m2v2 = 2.5 + 5v2$

By the law of Conservation Of Momentum, We know that.
$m1u1 + m2u2 = m1v1 + m2v2$
$0 = 2.5 + 5v2$
$- 2.5 = 5v2$
$v2 = - 0.5$
Therefore,
The velocity of the recoil of the gun is 0.5 m/s.

Thanks.

Answer 2

$\huge{\fcolorbox{red}{pink}{ Solution :)}}$

Given ,

• Mass of bullet = 10 g or 0.01 kg
• Mass of gun = 5.0 kg
• Velocity of bullet = 250 m/s

By using law of conservation of momentum ,

$\sf \large \fbox{Mass \: of \: gun × recoil \: velocity \: of \: gun = Mass \: of \: bullet × velocity \: of \: bullet }$

Substitute the known values , we get

5.0 × Recoil velocity of gun = 0.01 × 250

5.0 × Recoil velocity of gun = 2.5

Recoil velocity of gun = 0.5 m/s

Hence , the Recoil velocity of gun is 0.5 m/s

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