Question

A bullet of mass 18.8 grams is fired with a speed of 399 meters per second toward a wood block of mass 277 grams initially at rest on a very smooth surface. What is the change in momentum of the bullet if it becomes embedded in the block? Answer must be in 3 significant digits.

Answer 1

Given data:

* The mass of bullet is 18.8 grams. or 0.0188 kg.

* The speed of the bullet is 399 meter per second.

* The mass of the wood block is 277 grams ot 0.277 kg.

Solution:

The initial momentum of the syetm is,

`_{}p_i=m_1u_1+m_2u_{2\text{ }}`

where m_1 is the mass of bullet, m_2 is the mass of the wood block, u_1 is the initial velocity of the bullet, and u_2 is the initial velocity of the wood block,

Substituting the known values,

`\begin{gathered} p_i=0.0188*399+0 \\ p_i=7.5012kgms^(-1) \end{gathered}`

The final momentum of the system is,

`p_f=(m_1+m_2)v`

where v is the final velocity of the bullet and wooden block,

Substituting the known values,

`\begin{gathered} p_f=(0.0188+0.277)* v \\ p_f=0.2958v \end{gathered}`

By the law of conservation of momentum,

`\begin{gathered} p_i=p_f \\ 7.5012=0.2958v \\ v=(7.5012)/(0.2958) \\ v=25.36ms^(-1) \end{gathered}`

Thus, the change in the momentum of bullet is,

`\begin{gathered} p_f-p_i=25.36*0.0188-0.0188*399 \\ \Delta p=0.477-7.5012 \\ \Delta p=-7.02 \end{gathered}`

Here negative sign indicates the decrease in the momentum,

Thus, the change in the momentum of the bullet is 7.02 kg meter per second .