The expression for the bullet's speed vbullet is \(v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}\) and the speed of a 9.0 g bullet is v=16.8 m/s
Explanation:
A bullet of mass m is fired into a block of mass m that is at rest. the block, with the bullet embedded, slides distance d across a horizontal surface. the coefficient of kinetic friction is μk.
The initial kinetic energy of the bullet is given by
\(K_i = \frac{1}{2}mv^2\)
where
m is the mass of the bullet
v is the initial speed of the bullet
The expression for the bullet's speed vbullet is
\(F=\mu_k (m+M) g\)
where
\(\mu_k\) is the coefficient of kinetic friction
g is the acceleration of gravity
The work done by the force of friction is
\(W=-Fd = -\mu_k (m+M)g d\)
where d is the displacement of the block+bullet.
Because the final kinetic energy is zero (the bullet with the block comes at rest), we can write:
\(W=K_f - K_i = -K_i\)
And so
\(-\mu_k (m+M) g d = -\frac{1}{2}mv^2\)
By solving for v, the solution for the bullet speed:
\(-\mu_k (m+M) g d = -\frac{1}{2}mv^2\\v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}\)
The speed of a 9.0 g bullet that, when fired into a 12 kg stationary wood block causes the block to slide 5.4 cm across a wood table. Assume that k=0.20.
We have:
the mass of the bullet, m = 9.0 g = 0.009 kg
the mass of the block, M = 12 kg
the distance covered by the block+bullet, d = 5.4 cm = 0.054 m
the coefficient of friction, \(\mu_k = 0.20\)
the acceleration of gravity, \(g = 9.8 m/s^2\)
By substituting, we got
\(v=\sqrt{\frac{2 (0.20) (0.009 kg+12 kg)(9.8 m/s^2)(0.054 m)}{0.009 kg}}=16.8 m/s\)
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