A pistol of mass 2 kg fires a bullet of mass 50 g.
The bullet strikes a stationary block of mass
1
2 kg.
If the block, with the bullet embedded in it, moves
with a velocity of 4 m/s, the recoil velocity of the
pistol will be (Ignore friction and air resistance)
To answer this question, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the equation:
momentum = mass x velocity
Before the collision:
Momentum of the bullet = (mass of the bullet) x (velocity of the bullet)
= (0.050 kg) x (velocity of the bullet)
Momentum of the pistol = (mass of the pistol) x (velocity of the pistol)
= (2 kg) x (velocity of the pistol)
Initially, the block is stationary, so its momentum is zero.
After the collision:
The bullet and the block move together with a final velocity of 4 m/s. Therefore, the momentum of the bullet and the block combined is:
Final momentum of the bullet and block = (total mass of bullet and block) x (final velocity of bullet and block)
According to the conservation of momentum:
Initial momentum = Final momentum
Therefore, we can write the equation:
(mass of the bullet) x (velocity of the bullet) + (mass of the pistol) x (velocity of the pistol) = (total mass of bullet and block) x (final velocity of bullet and block)
Substituting the given values:
(0.050 kg) x (velocity of the bullet) + (2 kg) x (velocity of the pistol) = (2.05 kg) x (4 m/s)
Now, we can solve for the velocity of the pistol:
(0.050 kg x velocity of the bullet) + (2 kg x velocity of the pistol) = 8.2 kg·m/s
Rearranging the equation:
2 kg x velocity of the pistol = 8.2 kg·m/s - (0.050 kg x velocity of the bullet)
Dividing by 2 kg:
velocity of the pistol = (8.2 kg·m/s - (0.050 kg x velocity of the bullet)) / 2 kg
Simplifying:
velocity of the pistol = 4.1 m/s - (0.025 kg x velocity of the bullet)
Therefore, the recoil velocity of the pistol will be 4.1 m/s minus 0.025 times the velocity of the bullet.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision (pistol + bullet) is equal to the total momentum after the collision (block + embedded bullet + pistol).
Let's denote the recoil velocity of the pistol as v.
Before the collision:
- The initial momentum of the bullet is calculated as the product of its mass (converted to kg) and its initial velocity (which can be assumed as zero since it was fired from a stationary pistol). So, the initial momentum of the bullet is (0.05 kg) * 0 m/s = 0 kg*m/s.
- The initial momentum of the pistol is also zero since it is initially at rest.
After the collision:
- The final momentum of the block is calculated as the product of its mass (0.5 kg) and its final velocity (4 m/s). So, the final momentum of the block is (0.5 kg) * 4 m/s = 2 kg*m/s.
- The final momentum of the embedded bullet is the product of its mass (0.05 kg) and the final velocity of the block (4 m/s). So, the final momentum of the embedded bullet is (0.05 kg) * (4 m/s) = 0.2 kg*m/s.
- The final momentum of the pistol is the product of its mass (2 kg) and its recoil velocity (v). So, the final momentum of the pistol is (2 kg) * v.
Now, we can set up the equation using the conservation of momentum principle:
Initial momentum = Final momentum
0 + 0 = 2 + 0.2 + 2v
Simplifying the equation:
0 = 2 + 0.2 + 2v
-2.2 = 2v
v = -2.2 / 2
v = -1.1 m/s
Therefore, the recoil velocity of the pistol, in this case, is -1.1 m/s. The negative sign indicates the opposite direction of motion compared to the block's velocity. Keep in mind that the magnitude of the recoil velocity is 1.1 m/s.