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Arisa [49]
3 years ago
5

A shotgun of mass 3kg fires a bullet of 0.1kg at a velocity of 250m/s. What is the recoil velocity of the gun

Physics
1 answer:
alisha [4.7K]3 years ago
6 0

Answer:

8.33 m/s

Explanation:

The formula to apply is that of conservation of linear momentum taking that the this case is a single fire non reciprocating action.

Thus,

m₁v₁ = m₂v₂  where;

m₁ = mass of the shotgun = 3kg

v₁ = recoil velocity of the gun

m₂ = mass of bullet

v₂ = velocity of bullet

Substitute values in the equation as;

m₁v₁ = m₂v₂

3*v₁ =0.1*250

3v₁ =25

v₁ =25/3 = 8.33 m/s

<u>The recoil velocity of the gun is 8.33 m/s</u>

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Answer:

t = 2.5 seconds

Explanation:

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Explanation;

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3 years ago
Read 2 more answers
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nlexa [21]

Answer:

The given phenomenon  supports the Principle of Conservation of momentum.

Explanation:

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and we are given

m1 = mass of car = 600 kg

m2 = mass of van = 400 kg

u1 = Initial Speed of Car  

to find the initial speed we use equation of motion

Vf = Vi + at

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u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum  = (600 kg)(12 m/s) + (400 kg)(0 m/s)

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final momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = final speed of car+van (both locked ) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns   -------- (2)

on comparing (1) and (2)

Initial momentum = Final Momentum

Hence, the phenomenon of the system supports the principle of conservation of momentum.

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3 years ago
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statuscvo [17]

Answer:

ground

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4 years ago
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kompoz [17]

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