Math Problem Statement

A block with mass 3.96 kg is sitting at rest on a horizontal frictionless surface. A bullet with mass 0.0400 kg is traveling horizontally at 200 m/s. The bullet strikes the block and becomes embedded in it. If the coefficient of kinetic friction between the block and the surface is 0.400, what distance does the block slide after the collision before coming to rest?

Solution

Let's break down this problem step by step:

Given:

  • Mass of block: M=3.96 kgM = 3.96 \, \text{kg}
  • Mass of bullet: m=0.0400 kgm = 0.0400 \, \text{kg}
  • Initial velocity of bullet: vb=200 m/sv_b = 200 \, \text{m/s}
  • The block is initially at rest, so its initial velocity: vblock=0 m/sv_{\text{block}} = 0 \, \text{m/s}
  • Coefficient of kinetic friction: μk=0.400\mu_k = 0.400

We are to find the distance the block slides after the collision.

Step 1: Apply Conservation of Momentum

Since the block is at rest and the bullet embeds itself into the block, we have a perfectly inelastic collision.

The total momentum before collision is equal to the total momentum after collision: mvb=(M+m)vfinalm v_b = (M + m) v_{\text{final}}

Solving for vfinalv_{\text{final}} (velocity of the block and bullet after the collision):

= \frac{8}{4.00} = 2.00 \, \text{m/s}$$ ### Step 2: Work-Energy Principle Once the bullet and block are moving together, kinetic friction will cause the block to come to rest. The frictional force $$F_f$$ is given by: $$F_f = \mu_k (M + m) g$$ where $$g = 9.8 \, \text{m/s}^2$$ is the acceleration due to gravity. $$F_f = 0.400 \times (3.96 \, \text{kg} + 0.0400 \, \text{kg}) \times 9.8 \, \text{m/s}^2 = 0.400 \times 4.00 \times 9.8 = 15.68 \, \text{N}$$ Now, using the work-energy theorem, the work done by the friction force to stop the block is equal to the initial kinetic energy of the block. Kinetic energy of the block and bullet system: $$KE = \frac{1}{2} (M + m) v_{\text{final}}^2 = \frac{1}{2} (4.00 \, \text{kg}) (2.00 \, \text{m/s})^2 = 8.00 \, \text{J}$$ Work done by friction is $$W = F_f d$$, where $$d$$ is the distance traveled. Setting the work done by friction equal to the initial kinetic energy: $$F_f d = 8.00 \, \text{J}$$ $$15.68 \, \text{N} \times d = 8.00 \, \text{J}$$ $$d = \frac{8.00}{15.68} = 0.510 \, \text{m}$$ Thus, the distance the block slides before coming to rest is approximately **0.51 meters**. ### Correct Answer: (a) 0.51 m Would you like further details on any of the steps? Here are 5 related questions to deepen your understanding: 1. How does the coefficient of friction affect the distance the block slides? 2. What would happen if the collision were elastic instead of inelastic? 3. How is the work-energy principle applied in problems involving friction? 4. How would the result change if the bullet's velocity were doubled? 5. Why is momentum conserved in this inelastic collision? **Tip**: Always check for external forces when applying conservation of momentum to ensure they are negligible during the collision.

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Math Problem Analysis

Mathematical Concepts

Conservation of Momentum
Work-Energy Principle
Friction

Formulas

m v_b = (M + m) v_final
F_f = μ_k (M + m) g
Kinetic Energy = 1/2 (M + m) v_final^2
Work done by friction: W = F_f d

Theorems

Conservation of Linear Momentum
Work-Energy Theorem

Suitable Grade Level

College level Physics or advanced high school

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