91Ó°ÊÓ

The given data can be listed below as-

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

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.

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

$p=mv=mass×velocity$

Here, p is the total momentum of the of the lumps before collision, $p_1 and p_2$are the momentum of the first and the second lump respectively.

Substituting the values in the above equation, we get-

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

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

$(0,0.09,-0.18) kg.m/s$.

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

$(0,0.09,-0.18) kg.m/s$.

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 $m_1 and m_2$are the masses of the bodies.

Substituting the values in the above equation, we get-

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

Thus, the velocity of the stuck-together lump just after the collision is $(0,1.5,-3) m/s$.