Question

A ballistic pendulum is a device for measuring bullet speeds. One of the simplest versions consists of a block of wood hanging from two long cords. (Two cords are used so that the bottom face of the block remains parallel to the floor as the block swings upward.) A 0.013-kg bullet traveling at 320 m/s hits a 3.0-kg ballistic pendulum. However, the block is not thick enough for this bullet, and the bullet passes through the block, exiting with one-third of its original speed.

Required:
How high above its original position does the block rise?

Answer 1

Answer:

0.043 m upwards

Explanation:

The mass of the bullet, $m_1$ = 0.013 kg

Mass of the ballistic pendulum, $m_2$ = 3 kg

Velocity of the bullet, $v_1$ = 320 m/s

Therefore, from the law of conservation of momentum, we get

$m_1v_1+m_2v_2=m_1v_1^1+m_2v_2^1$

$(0.013 \times 320)+(3 \times 0) = \left(0.013 \times \frac{320}{0}\right) + (3 \times v_2^1)$

$3 \times v_2^1=2.774$

$v_2^1=0.92 \ m/s$

Therefore the required height to rise the block is given by :

$(v_2^1)^2-v_2^2=2gh$

$h=\frac{(v_2^1)^2-v_2^2}{2g}$

$h=\frac{(0.92)^2-0}{2(-9.81)}$

$h=-0.043 \ m$

Therefore, the block moves upwards for 0.043 meters.