Question

(15 points) A high-velocity 15-g bullet is shot vertically into a 2-kg block. The block lifts 0.8m straight upward. The bullet penetrates the block and becomes embedded in the block within a time interval of 0.0010 s. What is the velocity of the bullet when it hits the block?

Answer 1

Answer:

The bullet has a velocity of $\mathbf{32.47m/s}$ when it hits the block.

Explanation:

By energy conservation principle, we have

$\Delta E=0\\K_i+U_i=K_f+U_f\\mv^2=(m+M)gh\\v=\sqrt{\left(1+\frac{M}{m}\right)gh}=\sqrt{\left(1+\frac{2.0kg}{0.015kg}\right)\times 9.81\frac{m}{s^2}\times 0.8m} \approx \mathbf{32.47\frac{m}{s}}.$