Question

You and a friend each hold a lump of wet clay. Each lump has a mass of $\mathbf{30}\mathbf{g}\mathbf{.}$You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was $\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$, and the velocity of the other lump was$\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$. (a) What was the total momentum of the lumps just before the impact? (b) What is the momentum of the stuck-together lump just after the collision? (c) What is the velocity of the stuck-together lump just after the collision?

Short answer

a) the total momentum of the lumps just before the impact is $(0,0.09,-0.18)\hspace{0.33em}kg.m/s$, b) the momentum of the stuck-together lump just after the collision is $(0,0.09,-0.18)\hspace{0.33em}kg.m/s$and c) the velocity of the stuck-together lump just after the collision is$(0,1.5,-3)m/s.$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The lump has a mass of $0.03\hspace{0.33em}kg.$
• The velocity of one lump was$(3,3,-3)\hspace{0.33em}m/s$.
• The velocity of the other lump is $(-3,0,-3)\hspace{0.33em}m/s$.

Significance of the law of conservation of momentum for the lumps

This law states that the total momentum of an object before and after the collision becomes equal if no external forces are involved.

The equation of the law of conservation of momentum gives the momentum before and after the impact along with the velocity of the lump.

Determination of the momentum before and after the collision along with the velocity

a) From the law of conservation of momentum, the equation of the total momentum of the lumps before the impact is expressed as:

$p=p_1+p_2$

where

$p=mv=mass\times velocity$

Here, p is the total momentum of the of the lumps before collision, $p_1\hspace{0.33em}and\hspace{0.33em}{p}_2$are the momentum of the first and the second lump respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=(0.03\hspace{0.33em}kg)\times \left(\right(3,3,-3)\hspace{0.33em}m/s)+(0.03\hspace{0.33em}kg)\times \left(\right(-3,0,-3)\hspace{0.33em}m/s) \\ p=(0,0.09,-0.18)\hspace{0.33em}kg.m/s \end{gathered}$$

Thus, the total momentum of the lumps just before the impact is

$(0,0.09,-0.18)\hspace{0.33em}kg.m/s$.

b) According to the law of conservation of momentum, the total momentum before and after collision remains same.

Thus, the momentum of the stuck-together lump just after the collision is

$(0,0.09,-0.18)\hspace{0.33em}kg.m/s$.

c) According to the law of conservation of momentum, the velocity of the stuck-together lump just after the collision is expressed as-

$v=\frac{p}{m_1+m_2}$

Here, v is the velocity of the stuck-together lump, p is the momentum of the body and $m_1\hspace{0.33em}and\hspace{0.33em}{m}_2$are the masses of the bodies.

Substituting the values in the above equation, we get-

$$\begin{gathered} v=\frac{(0,0.09,-0.18)\mathrm{kg}·\mathrm{m}/\mathrm{s}}{0.03\mathrm{kg}+0.03\mathrm{kg}} \\ v=(0,1.5,-3)\hspace{0.33em}m/s \end{gathered}$$

Thus, the velocity of the stuck-together lump just after the collision is $(0,1.5,-3)\hspace{0.33em}m/s$.