Question

Car A of mass 1000 kg moving at 20m/s collides with a car B of mass 1200 kg moving at 10m/s in same direction. If the car B is shunted towards at 15m/s by the impact, what is the velocity, v, of the car A immediately after the crash ?

Answer 1

Answer on Question #59211, Physics / Mechanics | Relativity |

Car A of mass $1000\,\mathrm{kg}$ moving at $20\,\mathrm{m/s}$ collides with a car B of mass $1200\,\mathrm{kg}$ moving at $10\,\mathrm{m/s}$ in same direction. If the car B is shunted towards at $15\,\mathrm{m/s}$ by the impact, what is the velocity, $v$, of the car A immediately after the crash?

Solution:

Given:

$$\begin{array}{l} m_{1} = 1000\,\mathrm{kg}, \\ m_{2} = 1200\,\mathrm{kg}, \\ v_{1i} = 20\,\mathrm{m/s}, \\ v_{2i} = 10\,\mathrm{m/s}, \\ v_{2f} = 15\,\mathrm{m/s}, \\ v_{1f} = ? \end{array}$$

The equation that denotes the conservation of momentum is:

$$Momentum\ before\ collision = Momentum\ after\ collision$$ $$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$$

where, $m_{1} =$ mass of car A

$$\begin{array}{l} m_{2} = \text{mass of car B} \\ v_{1i} = \text{initial velocity of car A} \\ v_{2i} = \text{initial velocity of car B} \\ v_{2f} = \text{final velocity of car B} \end{array}$$

From above equation we have,

$$v_{1f} = \frac{m_{1}v_{1i} + m_{2}v_{2i} - m_{2}v_{2f}}{m_{1}}$$ $$v_{2f} = \frac{1000 \cdot 20 + 1200 \cdot 10 - 1200 \cdot 15}{1000} = 14\,\mathrm{m/s}$$

Answer: $14\,\mathrm{m/s}$

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