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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


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

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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