Question

A bullet with a mass of 0.04-kg is traveling horizontally at 1400 m/s when it strikes a stationary target that is balanced on the top of a wooden post. The target has a mass of 9.96-kg. The target, with the bullet embedded, is knocked off of the post and moves to the right. How fast is the target with the bullet lodged in it traveling?

Answer 1

The speed of the target with the bullet lodged in it is 5.6 m/s.

What is speed?

This is the rate of change of distance.

To calculate the speed of the target and the bullet lodged, we use the formula below.

Formula:

• V = mu/(m+M)............. Equation 1

Where:

• V = Speed of the target and the bullet
• m = mass of the bullet
• M = mass of the target
• u = Initial speed of the target

From the question,

Given:

• m = 0.04 kg
• u = 1400 m/s
• M = 9.96 kg

Substitute these values into equation 1

• V = (0.04×1400)/(0.04+9.96)
• V = 56/10
• V = 5.6 m/s

Hence, the speed of the target with the bullet lodged in it is 5.6 m/s.

Learn more about speed here: https://brainly.com/question/6504879

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