Question

A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450 N/m. The block and bullet continue to move, compressing the spring by 22.0 cm before the whole system momentarily comes to a stop. The coefficient of kinetic friction between block and the surface on which the block is resting is 0.35. Determine the initial speed of bullet.

Answer 1

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$.

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$

The initial speed of the bullet-block system is:

$v \approx 2.383\,\frac{m}{s}$

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

$(0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)$

$v_{o} = 889.199\,\frac{m}{s}$