A 1400 kg car moving south at 10.0 m/s collides with a 2350 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.36 m/s to the north. Find the velocity of the 2350 kg car before the collision.
m/s to the south
i have this same problem and im going to cry
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and its velocity (v).
Before the collision, we have the following momentum for each car:
- First car (1400 kg, moving south): p1 = (1400 kg) × (-10.0 m/s) = -14000 kg⋅m/s
- Second car (2350 kg, moving north): p2 = (2350 kg) × v2
After the collision, the cars stick together and move as a unit with a velocity of 5.36 m/s to the north:
- Velocity of the combination of the cars: v_final = 5.36 m/s
Applying the principle of conservation of momentum, we can equate the total momentum before the collision (p1 + p2) to the total momentum after the collision (p_final), and solve for v2:
p1 + p2 = p_final
-14000 kg⋅m/s + (2350 kg) × v2 = (3750 kg) × (5.36 m/s)
Simplifying the equation, we can solve for v2:
-14000 kg⋅m/s + (2350 kg) × v2 = 20100 kg⋅m/s
2350 kg⋅v2 = 34100 kg⋅m/s
v2 = 34100 kg⋅m/s / 2350 kg
v2 ≈ 14.51 m/s
Therefore, the velocity of the 2350 kg car before the collision is approximately 14.51 m/s to the south.