A pistol of mass 2 kg fires a bullet of mass 50g. The bullet strikes a stationary block of mass ½ kg. if the block, with the bullet in it, moves with a velocity of 4 m/s the recoil velocity of the pistol will be?

mass of block with bullet = .5 + .050 = .55 kg

momentum of block with bullet = .55*4
= 2.2 kg m/s

so momentum of bullet before it hits = 2.2

so .050 Vbullet = 2.2
Vbullet = 44m/s

but before we shot, the momentum was zero

0 = mass pistol * 0 + mass bullet * 0
no external foceres so after the shout momementum still 0

0 = 2 Vgun + momentum of bullet
= = 2 Vgun + 2.2
Vgun = - 1.1 m/s

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

The momentum before the bullet strikes the block is given by the sum of the momentum of the bullet and the pistol:

Initial momentum = (mass of bullet × velocity of bullet) + (mass of pistol × velocity of pistol)

Since the pistol is initially stationary, the velocity of the pistol is 0.

Initial momentum = (0.05 kg × velocity of bullet) + (2 kg × 0)

After the bullet strikes the block and they move together, their combined momentum is given by:

Final momentum = (mass of block + bullet) × velocity of block+bullet

Final momentum = (0.5 kg + 0.05 kg) × 4 m/s

According to the conservation of momentum, the initial momentum and final momentum must be equal:

Initial momentum = Final momentum

(0.05 kg × velocity of bullet) = (0.55 kg × 4 m/s)

Simplifying the equation:

0.05 kg × velocity of bullet = 2.2 kg m/s

Dividing both sides by 0.05 kg:

velocity of bullet = 2.2 kg m/s / 0.05 kg

velocity of bullet = 44 m/s

Therefore, the recoil velocity of the pistol will be 44 m/s in the opposite direction to the bullet's velocity.

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 the collision is equal to the total momentum after the collision.

Let's assume the recoil velocity of the pistol is v m/s.

Before the collision:
Momentum of the bullet = mass of the bullet * velocity of the bullet = (50g) * (0 m/s) = 0 kg.m/s
Momentum of the block = mass of the block * velocity of the block = (1/2 kg) * (0 m/s) = 0 kg.m/s
Momentum of the pistol = mass of the pistol * recoil velocity of the pistol = (2 kg) * (v m/s) = 2v kg.m/s

After the collision:
Momentum of the bullet = mass of the bullet * velocity of the bullet = (50g) * (4 m/s) = 0.2 kg.m/s
Momentum of the block = mass of the block * velocity of the block = (1/2 kg) * (4 m/s) = 2 kg.m/s
Momentum of the pistol = mass of the pistol * recoil velocity of the pistol = (2 kg) * (v m/s) = 2v kg.m/s

According to the principle of conservation of momentum:
Total momentum before the collision = Total momentum after the collision
0 kg.m/s + 0 kg.m/s + 2v kg.m/s = 0.2 kg.m/s + 2 kg.m/s + 2v kg.m/s

Simplifying the equation:
2v kg.m/s = 2.2 kg.m/s
v = 2.2 kg.m/s / 2 kg
v ≈ 1.1 m/s

Therefore, the recoil velocity of the pistol will be approximately 1.1 m/s.