a bullet of mass 20 gram is horizontly fired with a velocity 150 mtr/sec from a pistol of mass 2 kg. what is the recoil velocity of the pistol?
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Mass of bullet ( m1 ) = 0.02 kg ( after changing into gram to kilogram )
=> initial velocity ( u1 ) = 0 m/s
=> final velocity ( v1 ) = 150 m/s
Mass of pistol ( m2 ) = 2kg
=> initial velocity ( u2 ) = 0 m/s
=> final velocity ( v2 ) = ?
Let the final velocity of pistol be v
so,
BY THE LAW OF CONSERVATION OF MOMENTUM
=> m1u1 + m2u2 = m1v1 + m2v2
=> 0.02 × 0 + 2 × 0 = 0.02 × 150 + 2 × v
0 + 0 = 3 + 2v
3 = 2v
v = 3/2
v = 1.5 m/s
hope this will help you
0=m1•v1- m2•v2
v2 = m1•v1/ m2 =0.02•150/2 =1.5 m/s
Yes this question is right that's why I solve this question... 👍👍
To find the recoil velocity of the pistol, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided no external forces are acting on the system.
In this case, the initial total momentum of the system is zero because the pistol and bullet are initially at rest. After the bullet is fired, the bullet moves to the right with a certain velocity, and the pistol recoils to the left with a recoil velocity.
Let's denote the recoil velocity of the pistol as "V".
The initial momentum is given by:
Initial momentum = (Mass of the bullet × Velocity of the bullet) + (Mass of the pistol × Velocity of the pistol before firing)
Since the bullet was initially at rest, the velocity of the bullet before firing is 0. The equation becomes:
Initial momentum = 0 + (Mass of the pistol × Velocity of the pistol before firing)
The final momentum is given by:
Final momentum = (Mass of the bullet × Velocity of the bullet) + (Mass of the pistol × Velocity of the pistol after firing)
Now, we can equate the initial and final momenta:
0 + (Mass of the pistol × Velocity of the pistol before firing) = (Mass of the bullet × Velocity of the bullet) + (Mass of the pistol × Velocity of the pistol after firing)
Substituting the given values:
0 + (2 kg × 0 m/s) = (0.02 kg × 150 m/s) + (2 kg × V)
Simplifying the equation, we have:
0 = 3 kg⋅m/s + 2 kg⋅V
Rearranging the equation to isolate V (recoil velocity):
2 kg⋅V = -3 kg⋅m/s
V = (-3 kg⋅m/s) / (2 kg)
V = -1.5 m/s
Therefore, the recoil velocity of the pistol is -1.5 m/s. Note that the negative sign indicates the direction, which is opposite to the firing direction of the bullet.