A 6kg mass traveling at speed 10m/s strikes a stationarymass head-on, and the two masses stick together.

(a) What was the initial total kinetic energy?

(b) What is the final speed?

(c) What is the final total kinetic energy?

(d) What was the increase in internal energy of the two masses?

Mass of the first object is$m_1=6Kg$.

Initial speed of the first object Speed is 10m/s

Mass of the second object is$m_2=6kg$.

Collisions between two objects naturally fall under Newton's third law of motion. When two objects collide, forces of equal magnitude and opposite direction are applied to each object. Such pressures frequently result in one thing gaining momentum and gaining speed while the other object slows down (lose momentum).

There is a concept where the initial total kinetic energy of the objects with the same masses is equal to the initial kinetic energy of the first object.

$\mathit{K}{\mathit{E}}_{\mathbf{i}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{i}}^{\mathbf{2}}$

(a)

Substitute the value in the formula.

$$\begin{gathered} K{E}_i=\frac{1}{2}m{v}_i^2 \\ K{E}_i=\frac{1}{2}(6kg)(10m/s) \\ K{E}_i=300J \end{gathered}$$

Therefore, the initial total kinetic energy is 300 J.

(b)

Derive the formula for the velocity

The objects have the same masses, but the second object has an initial speed which is at rest.

$P_i=P_{1,f}+P_{2,f}$

With the inelastic collision, both of the objects will have the same speed since they also have the same mass $P_{1,f}=P_{2,f}$.

The equation can be rewritten as $P_i=2P_{2,f}$.

With this, break down the formula according to its mass and velocity. then derive to find the value of:

$\begin{array}{rcl}{P}_i& =& 2P_{2,f}\\ {\text{mvv}}_i& =& 2{\text{mv}}_{1,f}\\ v_{1,f}& =& \frac{v_i}{2}\end{array}$

Simply substitute the values to the derived formula.

$$\begin{gathered} v_{1,f}=\frac{v_i}{2} \\ {v}_{1,f}=\frac{10m/s}{2} \\ {v}_{1,f}=5m/s \end{gathered}$$

Therefore, the final speed is 5m/s.

(c)

The final total kinetic energy of the masses:

$\begin{array}{rcl}K{E}_f& =& K{E}_{1,f}+K{E}_{2,f}\\ & =& \frac{1}{2}{m}_1{v}_f^2+\frac{1}{2}{m}_2{v}_f^2\\ & =& \frac{1}{2}{v}_f^2\left(m_1+m_2\right)\end{array}$

Solve the Problem:

$\begin{array}{rcl}K{E}_f& =& \frac{1}{2}\times (5{)}^2\times (6+6)\\ & =& 150\text{J}\end{array}$

Therefore, the final total kinetic energy is 150J.

(d)

From part (a), total initial kinetic energy:$K{E}_i=300J$

From part (c), total final kinetic energy:$K{E}_i=150J$

So, the increase in internal energy of the two masses is

$\begin{array}{rcl}K{E}_i+K{E}_f& =& 300+150\\ & =& 450J\end{array}$

Therefore, the increase in internal energy of the two masses is 450J.