Question

A 2 kg bullet hits a 19.8 kg block of wood at rest. If the bullet sticks into the wood and sends the block flying at 10.8 m/s away, what was the velocity of the bullet? Round your answer to the nearest hundredth and don't forget your units.

Answer 1

By using conservation of momentum, the initial velocity of a 2 kg bullet that sticks into a 19.8 kg block of wood and sends it flying at 10.8 m/s was calculated to be 117.72 m/s, rounded to the nearest hundredth.

Step-by-step explanation:

To find the initial velocity of the bullet that hits a block of wood, we can use the law of conservation of momentum. The momentum before the collision (bullet's momentum) should equal the total momentum after the collision (combined bullet and block momentum).

The formula for conservation of momentum is:

m1 × v1 + m2 × v2 = (m1 + m2) × vfinal

Where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, and v2 is the initial velocity of the block (which is 0 since the block is at rest).

Given that m1 = 2 kg (the mass of the bullet), m2 = 19.8 kg (the mass of the block), and vfinal = 10.8 m/s (the velocity of the block and the bullet after the collision), we can rearrange the equation to solve for v1:

v1 = ((m1 + m2) × vfinal) / m1

Let's calculate:

v1 = ((2 kg + 19.8 kg) × 10.8 m/s) / 2 kg

v1 = (21.8 * 10.8) / 2

v1 = 235.44 / 2

v1 = 117.72 m/s

The initial velocity of the bullet was 117.72 m/s, rounded to the nearest hundredth.