Question

A bullet of mass 20g is formed a gun of mass 2kg moves with a velocity of 150m/s. What is the recoil velocity of the gun

Answer 1

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Explore the Concept:

Initially the bullet was inside the gun and they were in the state of rest. The velocity is zero for them. According to Law of conservation of momentum,

• Initial momentum = Final momentum.

Finally when the bullet was shot from the gun, the velocity of the bullet will result in some recoil velocity of gun.

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GiveN:

• Mass of bullet = 0.02 kg
• Speed of bullet = 150 m/s
• Mass of the gun = 2 kg

To FinD:

• Recoil velocity of the gun?

Step-wise-Step Explanation:

Initial momentum = 0 (Both are in rest)

Final momentum = m1v1 - m2v2

According to law of conservation of momentum:

⇒ Final momentum = Initial momentum

⇒ m1v1 - m2v2 = 0

⇒ (0.02)150 - 2(v2) = 0

⇒ 3 - 2(v2) = 0

⇒ v2 = 1.5 m/s

Hence, the recoil vel. of the gun is 1.5 m/s.

Answer 2

Mass of bullet in grams = 20g

Mass of bullet in kgs = (1 Kg =1000g)

So, 20g= 20/1000= 0.020kg

Velocity of bullet = 150m/s

Mass of gun = 2kg

Now, as before shooting, both were in rest,

So, their initial momentum will be = 0

We know, law of conservation of mass is:-

Final momentum = Initial momentum

So, m1v1 - m2v2 = 0

So, 0.02(150)- 2(v2)=0

So, 3-2(v2)=0

V2 = 1.5m/s

Therefore, the recoil velocity of gun is 1.5m/s

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