A bullet of mass 10g is horizontally fired with a velocity 100 m/s from a gun of mass 10 kg. The gun will move backward with linear momentum (1) 10 kg m/s (2) 1 kg m/s (3) 5 kg m/s (4) 2 kg m/s
Answers
Answer:
YOURE ANSWER IS OPTION 2. I HAVE WRITTEN THE ANSWER IN NEGATIVE DUE TO SIGN CONVENTION AS IT MOVE BACKWARD
Answer:
The correct option is (2) 1 kg m/s
linear momentum = 1 kg m/s
Explanation:
Law of conservation of momentum:
- When there are no external forces acting on a system of bodies, the total momentum always remains constant.
- the total momentum of the system before and after the collision are equal (constant)
- expression: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Before firing:
Initially, before firing there are no external forces acting on the bullet-gun system
so, u₁ and u₂ are equal to zero
∴ initial momentum = m₁u₁ + m₂u₂
= m₁ × 0 + m₂ × 0
= 0
After firing:
the bullet goes forward and the gun goes backward due to the opposite momentum acting on the gun by the bullet (gains some velocity)
Let,
mass of bullet m₁ = 10 g = 10/1000 kg = 0.01 kg
velocity of bullet v₁ = 100 m/s
mass of gun m₂ = 10 kg
velocity of gun v₂ = v m/s
∴ final momentum = m₁v₁ + m₂v₂
= 0.01 × 100 + 10 × v
= 1 + 10*v
from the law of conservation of momentum
initial momentum = final momentum
0 = 1 + 10*v
10*v = -1
v = -1/10
v = -0.1 m/s (negative sign indicates gun moves in backward direction)
∴ velocity of gun v₂ = v m/s = - 0.1 m/s
linear momentum of gun = mass × velocity
= 10 kg × ( -0.1 ) m/s
= -1 kg m/s
Hence, The gun will move backward with linear momentum = 1 kg m/s
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