# Cars Collide on a Hill, Conservation of Momentum

• Bitmap
In summary, the two vehicles involved in a head-on collision have initial velocities of 21m/s and -29m/s and are on an inclined road with a 10 degree angle. The mass of Vehicle B is 1.3 times larger than the mass of Vehicle A. After the collision, the vehicles stick together and friction and drag forces are negligible. To find the velocity of the two-vehicle system after the collision, the conservation of momentum equation is used and the values for mass and initial velocity are plugged in. Gravity force is also a factor to consider, but its impact over the duration of the collision is negligible. Therefore, calculating the impulse is not necessary.
Bitmap
Homework Statement
While driving on a road that is inclined at an angle of 10 degrees above the horizontal, Vehicle A and Vehicle B are in a head-on collision lasting for 0.071 sec Just before the collision, Vehicle A had a velocity of 21m/s and Vehicle B had a velocity of -29m/s as shown below.
The mass of Vehicle B is 1.3 times larger than the mass of Vehicle A. After the collision, the vehicles remain stuck together. You may assume that, as the collision took place, friction and drag forces are negligible in comparison to the force between the two vehicles.
What is the velocity of the two-vehicle system immediately after the collision?
Relevant Equations
Trig Functions (Sin, Cos, Tan)
Conservation of Momentum: M(a) + M(b) = (m(a) + m(b))*V(f)
QUESTION:
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For the purposes of this problem, we will define the direction of Vehicle A's initial velocity as the positive direction:

While driving on a road that is inclined at an angle of 10 degrees above the horizontal, Vehicle A and Vehicle B are in a head-on collision lasting for 0.071 sec Just before the collision, Vehicle A had a velocity of 21m/s and Vehicle B had a velocity of -29m/s as shown below.

The mass of Vehicle B is 1.3 times larger than the mass of Vehicle A. After the collision, the vehicles remain stuck together. You may assume that, as the collision took place, friction and drag forces are negligible in comparison to the force between the two vehicles.

What is the velocity of the two-vehicle system immediately after the collision?

EXPLANATION/APPROACH
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I know that the momentum is conserved so that the total momentum when the cars are together is equal to the momentum when the cars are separate. However, I am struggling to add Trig to this equation, as I don't know whether this will provide more or less momentum, and when I should apply the formulae to solve the problem.

I can do this problem without the incline:

Momentum of A = M(a)

Mass of A = m(a)

Final Velocity = V(f)

M(a) + M(b) = (m(a) + m(b))*V(f)

From this, I can plug in the given ratios/values to come out with the final velocity.

Bitmap said:
From this, I can plug in the given ratios/values to come out with the final velocity.
Yes, and that would be fine even on the slope if it were all instantaneous. But there is a time gap between before and after the collision.

Bitmap said:
You may assume that, as the collision took place, friction and drag forces are negligible in comparison to the force between the two vehicles.
It is important to read this passage carefully. Friction and drag are negligible. What forces does that leave that may not be negligible?

This thread gives the curious, but I'm sure false, impression, that the OP is laughing at the responses he's received!

jbriggs444 said:
It is important to read this passage carefully. Friction and drag are negligible. What forces does that leave that may not be negligible?
Gravity force? Maybe I can calculate the vectors using trig?

haruspex said:
Yes, and that would be fine even on the slope if it were all instantaneous. But there is a time gap between before and after the collision.
I would need impulse for this, right?

PeroK said:
This thread gives the curious, but I'm sure false, impression, that the OP is laughing at the responses he's received!
Definitely not. Thanks, everyone for the help!

PeroK
Bitmap said:
I would need impulse for this, right?
As I implied, you already have the after crash velocity were it not for gravity. What difference will gravity make over the duration of the collision?

## 1. How does the conservation of momentum apply to cars colliding on a hill?

According to the law of conservation of momentum, the total momentum of a closed system remains constant. In the case of cars colliding on a hill, the total momentum of the two cars before the collision is equal to the total momentum of the two cars after the collision.

## 2. What factors affect the conservation of momentum in a car collision on a hill?

The conservation of momentum in a car collision on a hill is affected by the mass and velocity of the cars involved. The steeper the hill, the greater the change in potential energy and the higher the final velocities of the cars after the collision.

## 3. Does the angle of the hill impact the conservation of momentum in a car collision?

Yes, the angle of the hill can impact the conservation of momentum in a car collision. A steeper hill will result in a greater change in potential energy and therefore a greater change in the final velocities of the cars after the collision.

## 4. Can the conservation of momentum be applied to more than two cars colliding on a hill?

Yes, the conservation of momentum can be applied to any number of cars colliding on a hill. As long as the system is closed and there are no external forces acting on the cars, the total momentum will remain constant.

## 5. How does the conservation of momentum explain the outcome of a car collision on a hill?

The conservation of momentum explains the outcome of a car collision on a hill by showing that the total momentum before the collision is equal to the total momentum after the collision. This means that the final velocities of the cars will depend on their masses and initial velocities, as well as the angle of the hill.

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