Question

a 1300kg car moving north at 25 m/sec collides with a 2100 kg car moving east at 17 m/sec they stick together in what direction and with what speed do they move after the collision

Answer 1

a $1300\mathrm{kg}$ car moving north at $25\mathrm{m / sec}$ collides with a $2100\mathrm{kg}$ car moving east at $17\mathrm{m / sec}$ they stick together in what direction and with what speed do they move after the collision

Solution

Momentum of first car:

$$p _ {1} = m _ {1} * v _ {1} = 1 3 0 0 * 2 5 = 3 2 5 0 0 m * k g / s$$

Momentum of second car:

$$p _ {1} = m _ {2} * v _ {2} = 2 1 0 0 * 1 7 = 3 5 7 0 0 m * k g / s$$

So, according to the sketch above:

$$p _ {r e s} = \sqrt {p _ {1} ^ {2} + p _ {2} ^ {2}} = \sqrt {3 2 5 0 0 ^ {2} + 3 5 7 0 0 ^ {2}} \approx 4 8 2 7 7 m * \frac {k g}{s}$$

Thus, speed of the collided cars is:

$$p _ {r e s} = (m _ {1} + m _ {2}) v _ {r e s}$$ $$v _ {r e s} = \frac {p _ {r e s}}{(m _ {1} + m _ {2})} = \frac {4 8 2 7 7}{(2 1 0 0 + 1 3 0 0)} \approx \mathbf {1 4 . 2 m / s}$$

Angle $\alpha$ can be calculated as:

$$\alpha = \arctan \frac {p _ {1}}{p _ {2}} = \arctan \frac {3 2 5 0 0}{3 5 7 0 0} \approx 4 2. 3 {}^ {\circ}$$

Answer: After collision cars will moved with the speed $14.2 \, \text{m/s}$ . $\alpha = 42.3{}^{\circ}$