Car A(1750kg) traveling due south and car B (1450kg)due east. They reach the same intersection at the same time and collide. The cars lock together and move off at 35.8km/h[E31.6degS]. What was the velocity of each car before the collided?
I tried this question using the law of momentum i keep getting 65km/h but the answer is supposed to be 34.3km/h for car a and for car b 67.3km/h. Is there some other method to do this?
They already tell you the original DIRECTION of each car's motion.
All of the combined momentum in the south direction comes from car A's initial momentum.
All of the combined momentum in the east direction comes from car B's initial momentum. Putting that in equation form:
1750*Va = 35.8*3200*sin31.6
1450*Vb = 35.8*3200*cos31.6
Va = 34.3 km/h etc
It was not necessary to convert velocity to m/s units, to get the answer in the same km/h units
To solve this problem, you can use the principle of conservation of momentum. Let's break down the steps to find the velocity of each car before the collision:
Step 1: Convert the final velocity of the cars after the collision to meters per second (m/s):
Given: Final velocity = 35.8 km/h [E31.6degS]
We need to convert this to m/s since the masses of the cars are given in kilograms.
35.8 km/h = 35.8 * 1000 m / (60 * 60) s ≈ 9.94 m/s
Step 2: Convert the final velocity into its horizontal and vertical components:
Given: Final velocity = 9.94 m/s [E31.6degS]
Vertical component = 9.94 m/s * sin(31.6 degrees) ≈ -5.20 m/s
Horizontal component = 9.94 m/s * cos(31.6 degrees) ≈ 8.61 m/s
Step 3: Apply the principle of conservation of momentum:
Before the collision, the total momentum in the x-direction is conserved and the total momentum in the y-direction is conserved separately.
Total momentum in the x-direction:
Car A: mass (mA) * velocity (vA) = 1750 kg * vX (to be found)
Car B: mass (mB) * velocity (vB) = 1450 kg * vY (to be found)
Total momentum in the y-direction:
Car A: mass (mA) * velocity (vA) = 1750 kg * vY (to be found)
Car B: mass (mB) * velocity (vB) = 1450 kg * (-vX) (to be found)
Step 4: Set up the equations for momentum conservation in the x and y directions:
In the x-direction:
mA * vX + mB * vY = 1750 kg * vX + 1450 kg * vY = (mA + mB) * Vxf (final velocity in the x-direction)
In the y-direction:
mA * vY + mB * (-vX) = 1750 kg * vY - 1450 kg * vX = (mA + mB) * VYf (final velocity in the y-direction)
Step 5: Solve the equations for the velocities of Car A (vX and vY) and Car B (vX and vY):
1750 kg * vX + 1450 kg * vY = 3200 kg * vXf
1750 kg * vY - 1450 kg * vX = 3200 kg * vYf
Divide the first equation by 1750 kg and the second equation by 1450 kg to simplify:
vX + (1450/1750) * vY = (3200/1750) * vXf
vY - (1450/1750) * vX = (3200/1750) * vYf
Simplify further by multiplying both equations by 1750:
1750 * vX + 1450 * vY = 3200 * vXf
1750 * vY - 1450 * vX = 3200 * vYf
Now, we have two equations with two unknowns. You can solve this system of equations using substitution or elimination method.
The solution should give you the velocities of Car A (vX and vY) and Car B (vX and vY) before the collision.
Note: Please recheck your calculations or refer to the given solution to see if any information was missed in the question or when calculating the final velocity.
To solve this problem, we can use the principles of conservation of momentum and vector addition. Here's how you can determine the velocity of each car before the collision:
1. Identify the given information:
- Mass of Car A (mA) = 1750 kg
- Mass of Car B (mB) = 1450 kg
- Velocity of both cars after the collision (v) = 35.8 km/h[E31.6°S]
2. Convert the given velocity to meters per second (m/s) to maintain consistency with the SI unit system:
- Velocity of both cars after the collision (v) = 35.8 km/h = 35.8 * (1000/3600) = 9.94 m/s [E31.6°S]
3. Split the given final velocity into its horizontal and vertical components:
- Vertical component: vSin(31.6°) = 9.94 * sin(31.6°) = 5.00 m/s [South]
- Horizontal component: vCos(31.6°) = 9.94 * cos(31.6°) = 8.48 m/s [East]
4. Apply the principle of conservation of momentum in the horizontal direction. Since the cars had different initial velocities, we consider the equation as follows:
- (mA * vA) + (mB * vB) = (mA + mB) * (vA' + vB')
5. Invent variables to represent the initial velocities of Car A and Car B. Let's assume:
- Initial velocity of Car A = vA
- Initial velocity of Car B = vB
6. Apply the principle of conservation of momentum in the vertical direction:
- In this case, since car A is traveling south and car B is traveling east, there is no change in the vertical component of momentum. Hence, we can write:
- mA * vA - mB * vB = 0
7. Solve the above two equations simultaneously. Let's start with the vertical equation:
- 1750 kg * vA - 1450 kg * vB = 0
- Rearranging the equation, we get: vA = (1450 kg * vB) / 1750 kg = 0.828 * vB
8. Substitute the derived value of vA into the horizontal equation:
- (1750 kg * vA) + (1450 kg * vB) = (1750 kg + 1450 kg) * 8.48 m/s
- 1750 kg * (0.828 * vB) + 1450 kg * vB = 3200 kg * 8.48 m/s
- 1449.6 * vB = 27136
- vB = 18.71 m/s
9. Substitute the value of vB back into the equation for vA:
- vA = 0.828 * 18.71 m/s = 15.50 m/s
10. Convert the final answers to kilometers per hour (km/h):
- vA = 15.50 m/s * (3600/1000) = 55.8 km/h
- vB = 18.71 m/s * (3600/1000) = 67.4 km/h
Therefore, the velocity of Car A before the collision was 55.8 km/h, and the velocity of Car B before the collision was 67.4 km/h.