A 1300-kg car moving north at 27.0 m/s is struck by a 2165-kg car moving east at 17.0 m/s. The cars are stuck together. How fast and in what direction do they move immediately after the collision?
Write two equations of conservation of momentum
m1u1 + m1u2 = m1v1 + m2v2
One in the i direction (West to East)
One in the j direction (South to North)
v1 = v2 in both cases (i and j) because they stick together afterwards.
For example:
Velocity of the 1300kg car is 0i + 27j and
the velocity of the 2165kg car is 17i + 0j
So, find i afterwards (call it x) and j afterwards (call it y) and the velocity afterwards of the combined cars is xi + yj. You can find the velocity and angle from that then.
edit: m1u1 + m2u2 = m1v1 + m2v2
I had a small error in the previous post with that formula.
To determine the speed and direction of the cars immediately after the collision, we need to apply the principles of conservation of momentum and vector addition.
First, let's calculate the initial momentum of each car:
Momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v).
For the 1300-kg car moving north at 27.0 m/s:
Initial momentum of the first car = mass × velocity = 1300 kg × 27.0 m/s = 35,100 kg·m/s (north).
For the 2165-kg car moving east at 17.0 m/s:
Initial momentum of the second car = mass × velocity = 2165 kg × 17.0 m/s = 36,905 kg·m/s (east).
Since momentum is conserved in a collision, the total momentum before the collision is equal to the total momentum after the collision. In other words:
Total initial momentum = Total final momentum
Let the final velocity of the stuck-together cars be Vf, where Vf is a vector quantity with both magnitude (speed) and direction.
To solve for the final velocity, we can break it down into its north and east components using vector addition.
Let's assume the final velocity is at an angle θ to the east direction. We can use trigonometry to relate the magnitude and direction of Vf to its north and east components.
The north component of Vf (Vfn) is given by:
Vfn = Vf × sin(θ)
The east component of Vf (Vfe) is given by:
Vfe = Vf × cos(θ)
Substituting these expressions into the conservation of momentum equation:
(mass of the first car × initial velocity of the first car) + (mass of the second car × initial velocity of the second car)
= (mass of the stuck-together cars × final velocity of the cars)
(1300 kg × 27.0 m/s) + (2165 kg × 17.0 m/s) = (1300 kg + 2165 kg) × Vf
35,100 kg·m/s (north) + 36,905 kg·m/s (east) = 3465 kg × Vf
Now, we can calculate the magnitude and direction of Vf:
Magnitude of Vf:
|Vf| = √[(Vfn)^2 + (Vfe)^2]
Direction of Vf (θ):
θ = atan(Vfn/Vfe)
Substituting the values:
|Vf| = √[(Vf × sin(θ))^2 + (Vf × cos(θ))^2]
θ = atan[(Vf × sin(θ))/(Vf × cos(θ))]
Solving these equations simultaneously will give us the magnitude and direction of the final velocity.