A bullet of mass 10g is fired at a speed of 200m/s from a gun of mass 2.0 kg. The recoil velocity of the gun is
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Answer:
Given Mass of bullet, m1 = 20 g = 0.02 kg
Mass of pistol, m2 = 2 kg
Initial velocity of the bullet (u1) and pistol, u2 = 0
Final velocity of the bullet, v1 = 150 ms-1
Let the recoil velocity of the pistol be v
The total momentum of the pistol and bullet is zero before the fire (because both are at rest)
The total momentum of the pistol and bullet after it is fired is
= (0.02 kg x 150 ms-1) + (2 kg x v ms-1)
On calculation, we get,
= (3 + 2v) kg ms-1
Total momentum after the fire = Total momentum before the fire
3 + 2v = 0
2v = -3
v = -3/2
v = -1.5 m/s
Hence, the recoil velocity of the pistol is 1.5 m/s
Explanation:
I know it's wrong answer but just (sad) ....
( ꈍ_ꈍ)
Given: A bullet of mass 10g is fired at a speed of 200m/s from a gun of mass 2.0 kg.
To find: Recoil velocity of the gun
Explanation: Let the mass of the bullet be m1 and the mass of the gun be m2.
m1= 10 g
= 0.01 kg (1 kg= 1000 g)
m2= 2 kg
Let the initial velocity of the bullet and gun be u1 and u2 respectively.
Let the final velocity of the bullet and gun be v1 and v2 respectively.
Since the gun and bullet both are at rest initially,
u1= 0 and u2=0
v1= 200 m/s
Here, initial momentum of the system
= Initial momentum of bullet+ Initial momentum of the gun
= m1 * u1 + m2 * u2
= 0.01 * 0 + 2 * 0
= 0
Final momentum of the system
= Final momentum of bullet+ Final momentum of the gun
= m1 * v1 + m2 * v2
= 0.01 * 200 + 2 * v2
= 2 + 2* v2
The momentum of a system always remains constant.
Therefore,
Initial momentum= Final momentum
=> 0 = 2 + 2 * v2
=> -2 = 2 * v2
=> -2/2 = v2
=> v2 = -1 m/s
The negative sign shows that the velocity of the gun is opposite to that of the velocity of the bullet. This is called the recoil velocity of the gun.
Therefore, the recoil velocity of the gun is 1 m/s.