Two cars collide at an intersection. Car A, with a mass of 2000kg, is going from west to east, while car B, of mass 1300kg, is going from north to south at 15.0m/s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0 degrees south of east from the point of impact.

Question

Two cars collide at an intersection. Car A, with a mass of 2000kg, is going from west to east, while car B, of mass 1300kg, is going from north to south at 15.0m/s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0 degrees south of east from the point of impact.

1.) How fast were the enmeshed cars moving just after the collision?

2.) How fast was car A going just before the collision?

Answer 1

To answer these questions, we can use the principles of momentum conservation. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the collision between the two cars can be considered as an isolated system.

Let's break down the steps to calculate the answers:

1.) To find the speed of the enmeshed cars just after the collision, we need to calculate their total momentum.
The momentum of an object is given by the product of its mass and velocity.

Let's denote the final velocity of the enmeshed cars as Vf and their total mass as M.
Since the cars move at an angle of 65.0 degrees south of east, we need to calculate their horizontal and vertical components of velocity.

The vertical component of velocity can be found using the trigonometric function sine:
Vf_y = Vf * sin(65.0°)

Similarly, the horizontal component of velocity can be found using the trigonometric function cosine:
Vf_x = Vf * cos(65.0°)

The momentum of the enmeshed cars in the vertical direction will be zero since they move horizontally after the collision. Therefore, to find Vf, we only need the horizontal component of velocity.

The total momentum of the enmeshed cars just after the collision will be the sum of the momenta of car A and car B just before the collision.

Momentum_A = mass_A * velocity_A
Momentum_B = mass_B * velocity_B

Applying the momentum conservation principle:
Momentum_A + Momentum_B = Momentum_enmeshed

Here, mass_A = 2000 kg, mass_B = 1300 kg, velocity_B = 15.0 m/s, and the resulting momentum's vertical component is 0. Therefore:

Momentum_A = 2000 kg * velocity_A
Momentum_B = 1300 kg * 15.0 m/s

Momentum_A + Momentum_B = (2000 kg * velocity_A) + (1300 kg * 15.0 m/s)

Since the total vertical momentum is zero, we can simplify the equation:
(2000 kg * velocity_A) + (1300 kg * 15.0 m/s) = 0

Solving for velocity_A gives:
velocity_A = - (1300 kg * 15.0 m/s) / 2000 kg

velocity_A ≈ -9.75 m/s (negative sign indicates the direction opposite to the initial velocity of car A)

2.) To find the speed of car A just before the collision, we can use the formula for momentum again:
Momentum_A = mass_A * velocity_A

Rearranging the equation yields:
velocity_A = Momentum_A / mass_A

Plugging in the known values:
velocity_A = Momentum_A / 2000 kg

But we already calculated Momentum_A in the previous step:
velocity_A = (2000 kg * velocity_A) / 2000 kg

Simplifying the equation, we find:
velocity_A = velocity_A

So, the speed of car A just before the collision is -9.75 m/s (approximately).