Physics bullet question

Question

A bullet has a mass of 20 grams and is traveling at 275 m/s. It passes completely through a wooden block and emerges with a velocity of 130 m/s. If the block is   1.2 kg in mass, what is the velocity of the block after the bullet passes through.

Answer 1

Momentum (p)

Mass (m)

Velocity (v)

p=mv

Due to the conservation of momentum we have:

p_{i =} p_{f +} p_block

Where ,

p_f is the final momentum of the bullet

pi is the initial momentum of the bullet

p_block is the momentum of the block

The bullet's mass is in grams, it should be in Kilograms.

20 g = 0.02 kg

Plugging in the values, we have

(0.02*275) = (0.02*130) + (1.2* v)

where v is the velocity of the block, so we solve for v

v = [ (0.02*275) - (0.02*130) ] / 1.2 = 2.41 m/s

Answer 2

Since there are no external forces acting on the moving bullet and block system, we can apply conservation of momentum.

Initially, only the bullet is moving, so we can calculate the initial momentum as follows:

p_i = 0.020 kg * 275 m/s = 5.5 kg m/s

after the collision the momentum of the system is split between the block and the bullet as follows

p_f = 0.020 * 130 m/s + 1.2 kg * V_b where V_b is the velocity of the block after the interaction. By setting the initial momentum to the final momentum, we obtain an equation in which the only unknown quantity is the velocity of the bullet.