Question

14. Two clay balls are moving toward each other along a common straight line. The one on the
left has mass 3 kg and velocity 4 m/s. The one on the left has mass 2 kg and velocity 9 m/s. The
balls collide and form a single ball of mass 5 kg and keep moving. Some kinetic energy is lost as
heat in this process. Calculate this lost energy.

Answer 1

Answer on Question #42035 – Physics - Mechanics | Kinematics | Dynamics

14. Two clay balls are moving toward each other along a common straight line. The one on the left has mass $3\,\mathrm{kg}$ and velocity $4\,\mathrm{m/s}$. The one on the left has mass $2\,\mathrm{kg}$ and velocity $9\,\mathrm{m/s}$. The balls collide and form a single ball of mass $5\,\mathrm{kg}$ and keep moving. Some kinetic energy is lost as heat in this process. Calculate this lost energy.

Solution:

$$\begin{array}{l} \mathrm{m}_1 = 3\,\mathrm{kg} - \text{mass of the first ball}; \\ \mathrm{v}_1 = 4\,\frac{\mathrm{m}}{\mathrm{s}} - \text{initial velocity of the first ball}; \\ \mathrm{m}_2 = 2\,\mathrm{kg} - \text{mass of the second ball}; \\ \mathrm{v}_2 = 9\,\frac{\mathrm{m}}{\mathrm{s}} - \text{initial velocity of the second ball}; \\ \mathrm{u} - \text{velocity of the balls after collision} \\ E_{\text{heat}} - \text{lost kinetic energy as heat}; \end{array}$$

Law of momentum conservation along X-axis:

$$\begin{array}{l} \mathrm{m}_1 \mathrm{v}_1 - \mathrm{m}_2 \mathrm{v}_2 = (\mathrm{m}_1 + \mathrm{m}_2) \mathrm{u} \\ \mathrm{u} = \frac{\mathrm{m}_1 \mathrm{v}_1 - \mathrm{m}_2 \mathrm{v}_2}{\mathrm{m}_1 + \mathrm{m}_2} \end{array}$$

Law of energy conservation:

$$\begin{array}{l} E_{\text{initial}} = E_{\text{final}} \\ \mathrm{K E}_1 + \mathrm{K E}_2 = \mathrm{K E}_{1+2} + E_{\text{heat}} \\ \frac{\mathrm{m}_1 \mathrm{v}_1^2}{2} + \frac{\mathrm{m}_2 \mathrm{v}_2^2}{2} = \frac{(\mathrm{m}_1 + \mathrm{m}_2) \mathrm{u}^2}{2} + E_{\text{heat}} \\ E_{\text{heat}} = \frac{\mathrm{m}_1 \mathrm{v}_1^2 + \mathrm{m}_2 \mathrm{v}_2^2}{2} - \frac{(\mathrm{m}_1 + \mathrm{m}_2) \mathrm{u}^2}{2} \\ \text{(1)in(2):} \\ E_{\text{heat}} = \frac{\mathrm{m}_1 \mathrm{v}_1^2 + \mathrm{m}_2 \mathrm{v}_2^2}{2} - \frac{(\mathrm{m}_1 + \mathrm{m}_2)}{2} \frac{(\mathrm{m}_1 \mathrm{v}_1 - \mathrm{m}_2 \mathrm{v}_2)^2}{(\mathrm{m}_1 + \mathrm{m}_2)^2} = \\ = \frac{(\mathrm{m}_1 \mathrm{v}_1^2 + \mathrm{m}_2 \mathrm{v}_2^2)(\mathrm{m}_1 + \mathrm{m}_2) - (\mathrm{m}_1 \mathrm{v}_1 - \mathrm{m}_2 \mathrm{v}_2)^2}{2} = \end{array}$$ $$= \frac {\left(3 \mathrm {k g} \cdot \left(4 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2} + 2 \mathrm {k g} \cdot \left(9 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}\right) \left(3 \mathrm {k g} + 2 \mathrm {k g}\right) - \left(3 \mathrm {k g} \cdot 4 \frac {\mathrm {m}}{\mathrm {s}} - 2 \mathrm {k g} \cdot 9 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}}{2} = 5 0 7 \mathrm {J}$$

Answer: lost energy is equal to 507J.

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