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A 14 g bullet is fired into a 120 g wooden block initially at rest on a horizontal surface. The acceleration of gravity is 9.8 m/s^2. After impact, the block slides 6.97 m before coming to rest.

If the coefficient of friction between block and surface is 0.568, what was the speed of the bullet immediately before impact? Answer in units of m/s

I have tried doing this so far:

  1. Find Normal force $N = (m_1+m_2)g$
  2. Find $F_f = \mu N$
  3. Find $A$: $F_{net}=mA$
  4. Find speed: $v_f^2 = v_i^2 + 2ad$
  5. Find $v_i$: $m_1v_i=(m_1+m_2)v$

My answer was in the range of 740 m/s. I am confused.

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closed as too localized by David Z Dec 6 '11 at 8:42

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Sorry for all the edits if someone was reading it. I kept having errors in my typing! –  Clint Dec 4 '11 at 21:55
    
Bump* Can anyone help? –  Clint Dec 4 '11 at 23:22
    
Hi Clint, and welcome to Physics Stack Exchange! This isn't a site for homework help, but rather a site for general conceptual questions about physics. In other words, we prefer that you ask "what does this mean?" rather than "what am I doing wrong?" If you can edit your question to focus on a conceptual issue, rather than just asking for someone to check your work, I'll be happy to reopen it. See our homework policy for more info. –  David Z Dec 6 '11 at 8:46

2 Answers 2

Let $M$ be the mass of the block and $m$ be the mass of the bullet.

Let $v_{0}$ be the velocity of the bullet before the collision and $v$ be the velocity of the combined mass i.e after the collision.

From the Law of Conservation of Momentum,

$m \times v_0 + M \times 0 = \left(M +m\right)v$

$ v_0 = \frac{\left(M+m\right)}{m}v$

Now, After the Collision the Combined Mass (i.e the gun and the bullet) is acted upon by the frictional force, $f$

$f = \mu \times N \implies f = \mu \times (M +m)g $

From the Work-Energy Theorem,

$\frac{1}{2}(M+m)v^2 = f\times s$

$\frac{1}{2}(M+m)v^2 = \mu (M+m)g \times s$

$v = \sqrt{2 \mu gs}$

$v_0 = \frac{(M+m)}{m} \sqrt{2 \mu gs}$

Now After Calculating $v_0 = 84.31 ms^{-1}$ is what I get.

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Hi, and welcome to Physics Stack Exchange! This is a good answer, but our homework policy states that questions which give away complete answers to homework-like questions will be temporarily deleted, so accordingly I'm removing this for now. I'll undelete it after a couple of days. –  David Z Dec 6 '11 at 8:44

Your way to solve this problem is basically right, check your calculations.

The total friction force is the normal force times the friction coefficient.

$f = N\cdot\mu = (m_1+m_2)g\mu$

where $m_1$ is the mass of bullet, $m_2$ is the mass of block. From the conservation of energy, the work done by the friction should equal to the change of kinetic energy.

$f\cdot s = \frac{1}{2}(m_1+m_2)v^2$

where $v$ is the velocity after impact. Since the momentum should be conserved during this impact, so

$m_1v_{initial} + m_2\cdot 0 = (m_1+m_2)v$

So solve these three equations, you will get

$v_{initial} = \frac{m_1+m_2}{m_1}\sqrt{2g\mu s}$

Calculate it, you will get the velocity of bullet immediately before impact.

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