Question

Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed of 25 m/s before the collision. Car B has a mass of 1600 kg. The skid marks show that, immediately after the collision, the wreckage was moving in a direction making an angle of 40° with the original direction of car A. What was the speed of car B before the collision, assuming that any other unbalanced forces are negligible?

Answer 1

Answer:

$v_{B0}=15.73 m/s$

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

$p_{0x}=p_{fx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}$  

$m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40)$ (1)

y-coordinate

$p_{0y}=p_{fy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}$  

$m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40)$ (2)

We can divide equations (2) and (1):

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}$

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)$

$v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)$

$v_{B0}=\frac{1200*25}{1600}*tan(40)$

$v_{B0}=15.73 m/s$

I hope it helps you!

           

Answer 2

Answer:

The speed of car b before collision is 15.72 m/s

Explanation:

Given that,

Mass of car a= 1200 kg

Speed = 25 m/s

Mass of car b = 1600 kg

moving in a direction making an angle of 40°

We need to calculate the speed of car b before the collision

Using conservation of momentum

Horizontal momentum,

$P_{i}=P_{f}m_{1}v_{1}=(m_{1}+m_{2})\times v\cos\theta$

substitute the value into the formula

$1200\times25=(1200+1600)v\times\cos40v=\dfrac{1200\times25}{2800\times\cos40}v=13.98\ m/s$

Vertical momentum,

$P_{i}=P_{f}m_{2}v_{2}=(m_{1}+m_{2})\times v\sin\theta$

substitute the value into the formula

$1600\times v=(1200+1600)\times13.98\times\sin40\\v_{2}=\dfrac{2800\times13.98\sin40}{1600}\\v_{2}=15.72\ m/s$

Thus, The speed of car b before collision is 15.72 m/s