Question

To determine the muzzle velocity of a bullet red from a rifle, you shoot the 2-g bullet into a 2-kg wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of 0.5 cm above its initial position.
What is the velocity of the bullet on leaving the gun's barrel?

Answer 1

Answer:

313.3 m/s

Explanation:

$m$ = mass of the bullet = 2 g = 0.002 kg

$M$ = mass of the wooden block = 2 kg

$v$ = speed of the bullet before collision with block

$V$ = Speed of bullet-block combination after collision

$h$ = height gained by the combination after collision = 0.5 cm = 0.005 m

Using conservation of energy

Kinetic energy lost by the combination = Potential energy gained

$(0.5) (m + M) V^{2} = (m + M) gh\\V = \sqrt{2gh}\\V = \sqrt{2(9.8)(0.005)}\\V = 0.313 ms^{-1}$

Consider the collision between bullet and block

Using conservation of momentum

$m v = (m + M) V\\(0.002) v = (0.002 + 2) (0.313)\\v = 313.3 ms^{-1}$

Answer 2

The velocity of the bullet depends on the displacement. The velocity of the bullet on leaving the gun's barrel is 313.313 m/s.

What is velocity?

The velocity is defined as the displacement or change in position of the object per unit of time.

Given that the mass m of the bullet is 0.002 kg, the mass M of the wooden block is 2 kg. After the bullet is embedded in the block, the block swings up to a maximum height h of 0.5 cm above its initial position

Let's consider that v is the speed of the bullet before embedded with block and V is the speed of bullet block combination.

According to the law of energy conservation, the potential energy and kinetic energy will be equivalent.

$\dfrac {1}{2} (m+M)V^2 = (m+M)gh$

$V^2 = 2gh$

$V = \sqrt{2\times 9.8 \times 0.005}$

$V = 0.313 \;\rm m/s$

The momentum of the bullet will be equivalent to the momentum of the bullet block system.

$mv = (m+M)V$

$0.002 v= ( 0.002 + 2) 0.313$

$v = 313.313 \;\rm m/s$

Hence we can conclude that the velocity of the bullet on leaving the gun's barrel is 313.313 m/s.

To know more about the velocity, follow the link given below.

https://brainly.com/question/862972.