Question

a bullet of mass 20 gram is horizontally fired with a velocity 200 ms -1 from a pistol of mass 3 kg what is the recoil velocity of the pistol​

Answer 1

Answer:

ans

Explanation:

movementum of bullet is =movementum of gun

20/1000×200=3×v

4=3×v

v=4/3

v=1.3333

Answer 2

From question, we have;

• Mass of bullet = 20 g = 0.02 kg
• Velocity of bullet after firing = 200 m/s
• Mass of pistol = 3kg
• Initially, the the pistol was at rest and thus the intial velocity of pistol and bullet is 0.

Now, we know that;

According to law of conservation of momentum:

• Final momentum = Initial momentum
• $\sf{{m}_{1} {v}_{1} +{m}_{2} {v}_{2} = {m}_{1} {u}_{1} + {m}_{2} {u}_{2}}$

Now, just put all the values in the formula;

= $\sf{{m}_{1} {v}_{1} +{m}_{2} {v}_{2} = {m}_{1} {u}_{1} + {m}_{2} {u}_{2}}$

= (0.02×200) + (3×v) = (0.02×0) + (3×0)

= 4 + 3v = 0

• Now after firing, momentum of pistol(recoil) = momentum of bullet(firing)

= 3v = 4

= v = 4/3

= v = 1.33.. m/s

Therefore, the recoil velocity of the pistol will be 1.33 m/s.

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Thank youu!! :))