Question

In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block

Answer 1

Answer:

• Here we use the conservation of momentum theorem.

• m stands for mass, and v stands for velocity. The numbers refer to the respective objects.

• m1v1 + m2v2 = m1vf1 + m2vf2

• Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.

• m1v1 + m2v2 = vf(m1 + m2)

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s