Question

A gun is fired vertically into a 1.40 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 210 m/s, how high will the block rise into the air after the bullet becomes embedded in it?

Answer 1

To solve this problem we will use the theorems related to the conservation of momentum to determine the final speed of the system, once the bullet hits it. From there, we will calculate with the kinematic equations of linear motion, the height traveled with that speed. The conservation of momentum say us that,

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

Where,

$m_1 =$ mass of bullet

$m_2 =$ mass of block of wood

$u_{1,2} =$ Initital velocity of bullet and block of wood respectively.

$v_f =$ Final velocity

Our values are given as,

$m_1 = 1.40 Kg\\m_2 = 0.021 Kg\\u = 210 m/s$

Replacing we have that

$(1.40 + 0.021) * v_f = 0.021 * 210$

$v_f = 3.103 m/s$

Note that the initial velocity of wood block is 0 because is at rest

Now using the kinematic equation of motion we have that

$v_f^2-v_i^2 = 2gh$

As $v_i = 0$

$h = \frac{v_f^2}{2g}$

$h = \frac{3.103^2}{2 * 9.8}$

$h = 0.49 m$

Therefore the height reached by block after bullet embedding in it is 0.49 m