A 4 gram bullet is fired horizontally with a speed of 300m/s into a 0.8kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3, how far will the block slide? What fraction of the bullet's energy is dissipated in the collision itself?

Question

anonymous · Answer

F=ma so, force of bullet is F=4x10^-3x300=1.2N
I don't remember the formula for coefficient of friction... so wait a bit...

aaronq · Answer

you can use the momentum because it's conserved (you could also use the kinetic energy but i'm gonna go with the former)

let $m_1$ and$v_1$ be the mass and velocity of the bullet
let $m_2$ and$v_2$ be the mass and velocity of the block

$(m_1v_1)_f+(m_2v_2)_f=(m_1v_1)_i+(m_2v_2)_i$

The block is at rest initially, so $(v_2)_i=0$
It's an inelastic collision so $(v_1)_f=0$

so the formula reduces to: $(m_2v_2)_f=(m_1v_1)_i$

Solve for $(v_2)_f$:   $(v_2)_f=\dfrac{(m_1v_1)_i}{m_2}$

Now we need use newton's laws to find the acceleration of the block.

$\Sigma F_y=n-mg=0\rightarrow n=mg$

Friction is  $f=\mu n\rightarrow f=\mu mg$

$\Sigma F_x=-f=ma\rightarrow a=\dfrac{-f}{m}$

Therefore the acceleration is  $a=\dfrac{-\mu mg }{m}=-\mu g$

Now we use kinematics to find the distance travelled:

$v_f^2=v_i^2+2a\Delta x\rightarrow \Delta x=\dfrac{v_f^2-v_i^2}{2a}$

When the block stops, it's velocity is zero, $v_f=0$, the distance travelled is given by:

$\Delta x=\dfrac{0^2-(\dfrac{(m_1v_1)_i}{m_2})^2}{2(-\mu g)}=\dfrac{(\dfrac{(m_1v_1)_i}{m_2})^2}{2(\mu g)}$

somy · Answer

OMG great job!! @aaronq