Question

On a vehicle crash simulation, a car of mass 1200 kg is travelling at a velocity of 40 mph in the easterly direction when it is hit by a truck of mass 3300 kg travelling at a velocity of 25 mph from the westerly direction. Assuming that the two vehicles become entangled calculate in m/s their combined velocity after the crash.
Take 1.61 km/hr = 1 mph

Answer 1

Answer on Question #44566 – Physics – Acoustics

Question.

On a vehicle crash simulation, a car of mass $1200\,\mathrm{kg}$ is travelling at a velocity of 40 mph in the easterly direction when it is hit by a truck of mass 3300 kg travelling at a velocity of 25 mph from the westerly direction. Assuming that the two vehicles become entangled calculate in m/s their combined velocity after the crash.

Take $1.61\,\mathrm{km/hr} = 1\,\mathrm{mph}$

$$1\,\mathrm{mph} = 1.61\,\frac{\mathrm{km}}{\mathrm{hr}} = 0.447\,\frac{\mathrm{m}}{\mathrm{s}}$$

Given:

$$\begin{array}{l} m_1 = 1200\,\mathrm{kg} \\ v_1 = 40\,\mathrm{mph} \\ m_2 = 3300\,\mathrm{kg} \\ v_2 = 25\,\mathrm{mph} \\ \end{array}$$

Find:

$$v = ?$$

Solution.

Let us the law of conservation of momentum for this inelastic collision. Therefore,

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$$

So,

$$v = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

Calculate:

$$v = \frac{1200 \cdot 40 + 3300 \cdot 25}{1200 + 3300} = \frac{48000 + 82500}{4500} = 29\,\mathrm{mph} = 13\,\frac{\mathrm{m}}{\mathrm{s}}$$

If we will use the law of conservation of energy, we will receive the same answer:

$$\frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} = \frac{(m_1 + m_2) v^2}{2}$$ $$v = \sqrt {\frac {m _ {1} v _ {1} ^ {2} + m _ {2} v _ {2} ^ {2}}{m _ {1} + m _ {2}}}$$

Calculate:

$$v = \sqrt {\frac {1200 \cdot 40 ^ {2} + 3300 \cdot 25 ^ {2}}{1200 + 3300}} = 29 \text{ mph} = 13 \frac {m}{s}$$

Answer.

$$v = 13 \frac {m}{s}$$

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