Question

A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum consists of a stationary 2.50- kg block of wood suspended by a wire of negligible mass. A 0.0100- kg bullet is fired into the block, and the block (with the bullet in it) swings to a maximum height of 0.650 m above the initial position (see class note for drawing). Find the speed with which the bullet is fired, assuming that air resistance is negligible.

Answer 1

Answer:

$v_1 = 896.35 m/s$

Explanation:

As we know that bullet + pendulum system will move to the height of 0.650 m above the initial position

so here we can use energy conservation to find the speed just after the bullet hit the block

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.650)}$

$v = 3.57 m/s$

Now we can use momentum conservation to find the initial speed of the bullet

$m_1v_1 = (m_1 + m_2)v$

$0.0100 v_1 = (2.50 + 0.01)(3.57)$

$v_1 = 896.35 m/s$