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?

To find the recoil velocity of the pistol, we can use the principle of conservation of momentum.

The momentum before firing can be calculated using the formula:

Initial momentum of the bullet + Initial momentum of the pistol = 0

The momentum of an object can be calculated by multiplying its mass by its velocity.

Given:
Mass of the bullet (m1) = 20 grams = 0.02 kg
Velocity of the bullet (v1) = 150 m/s

Mass of the pistol (m2) = 2 kg
Velocity of the pistol (v2) = ?

Using the equation for conservation of momentum:

(mass of the bullet x velocity of the bullet) + (mass of the pistol x velocity of the pistol) = 0

(0.02 kg x 150 m/s) + (2 kg x velocity of the pistol) = 0

Simplifying the equation:

0.03 kgm/s + 2 kg x velocity of the pistol = 0

2 kg x velocity of the pistol = -0.03 kgm/s

Dividing both sides of the equation by 2 kg:

Velocity of the pistol = -0.03 kgm/s / 2 kg

Velocity of the pistol = -0.015 m/s

Therefore, the recoil velocity of the pistol is -0.015 m/s (negative sign indicates that the pistol moves in the opposite direction of the bullet's motion).

To find the recoil velocity of the pistol, we can use the principle of conservation of momentum. This principle states that the total momentum of an isolated system remains constant before and after an event.

In this case, the bullet and the pistol are part of an isolated system. Before the firing, the system is at rest, so the initial momentum is zero. After the firing, the bullet moves in one direction, and the pistol moves in the opposite direction to conserve momentum.

According to the conservation of momentum:

m_b * v_b + m_p * v_p = 0

Where:
m_b is the mass of the bullet
v_b is the velocity of the bullet
m_p is the mass of the pistol
v_p is the velocity of the pistol

We have the following values:
m_b = 20 grams = 0.02 kg
v_b = 150 m/s
m_p = 2 kg
v_p = ?

Now we can solve the equation for v_p:

(0.02 kg * 150 m/s) + (2 kg * v_p) = 0

(0.02 kg * 150 m/s) = - (2 kg * v_p) (rearranging the equation)

- 3 kg*m/s = 2 kg * v_p

v_p = - (3 kg*m/s) / 2 kg

v_p = -1.5 m/s

Therefore, the recoil velocity of the pistol is -1.5 m/s, indicating that it moves in the opposite direction to the bullet. The negative sign indicates the opposite direction of motion compared to the bullet's positive velocity.

balance momentum:

.020 * 150 = 2v

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