Question

A 1720 kg car skidding due north on a level frictionless icy road at 239.44 km/h collides with a 2597.2 kg car skidding due east at 164 km/h in such a way that the two cars stick together. 2597.2 kg 164 km/h 239.44 km/h 1720 kg vf θ N At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at? Answer in units of ◦ .

Answer 1

Answer:

$\theta=44.03^{o}$

Explanation:  

Here we have an inelastic collision problem. We can use the momentum (p = mv) conservation law in each component of the displacement.

So, $p_{i}=p_{f}$

X-component:

$m_{1}v_{i1x}+m_{2}v_{i2x}=m_{1}v_{f1x}+m_{2}v_{f2x}$ (1)

Now,

• v(i1x) is 0, because the first car just moving in y-direction
• v(i2x) is 164 km/h
• v(f1x)=v(f2x), because both cars stick together after the collision, so they have the same x-component velocity.

Then, using this information we can rewrite the equation (1).

$m_{2}v_{i2x}=v_{fx}(m_{1}+m_{2})$

$v_{fx}=\frac{m_{2}v_{i2x}}{m_{1}+m_{2}}=\frac{2597.2*164}{1720+2597.2}$

$v_{fx}=98.66 km/h$

Y-component:

$m_{1}v_{i1y}+m_{2}v_{i2y}=m_{1}v_{f1y}+m_{2}v_{f2y}$ (2)

We can do the same but with the next conditions:

• v(i1y) is 239.44 km/h
• v(i2y) is 0, because the second car just moving at the x-direction
• v(f1y)=v(f2y), because both cars stick together after the collision, so they have the same y-component velocity.

Then, using this information we can rewrite the equation (2).

$m_{1}v_{i1y}=v_{fy}(m_{1}+m_{2})$

$v_{fy}=\frac{m_{1}v_{i1y}}{m_{1}+m_{2}}=\frac{1720*239.44}{1720+2597.2}$

$v_{fy}=95.39 km/h$

Now, as we have both components of the final velocity, we can find the angle East of North. Using trigonometric functions, we have:

$tan(\theta)=\frac{v_{y}}{v_{x}}$

$\theta=arctan(\frac{v_{y}}{v_{x}})=arctan(\frac{95.39}{98.66})$

$\theta=44.03^{o}$

I hope it helps you!