Question

A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .

Answer 1

The body moves at a velocity of 1.62m/s after the bullet emerges.

Given:

Mass of bullet, $m_1$ = 22g

                               = 0.022 kg

Mass of the block, $m_2$ = 1.9 kg

Velocity of bullet , $v_1$ = 265 m/s

$v_2 = 0$

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

$v^{'}_1 =125 m/s$

$v^{'}_2=?$

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

   $m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s$

Therefore, It moves with a velocity of 1.62 m/s.

Learn more about momentum here:

brainly.com/question/25121535

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