A 2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 39.0° north of east and at a speed of 5.15 m/s. Find the speed of the 3,000-kg car before the collision.
keep the momentum the same
2000*10i + 3000vj = 5000*5.15(cos39°i + sin39°j)
Now, 5000*5.15*cos39° = 20000
so, 3000v = 5000*5.15*sin39° = 16196
v = 5.40
To find the speed of the 3,000-kg car before the collision, we can use the principle of conservation of momentum.
Let's start by breaking down the given information:
Mass of the first car, m1 = 2,000 kg
Initial velocity of the first car, v1i (moving east) = 10.0 m/s
Mass of the second car, m2 = 3,000 kg
Initial velocity of the second car, v2i (moving north) = unknown
After the collision:
Velocity of the combined cars, vf (moving at an angle of 39.0° north of east) = 5.15 m/s
According to the conservation of momentum, the sum of the momenta of the two cars before the collision should be equal to the sum of their momenta after the collision.
The momentum of an object is given by the product of its mass and velocity:
Momentum (p) = mass (m) × velocity (v)
Using the principle of conservation of momentum, we can express this as an equation:
m1 × v1i + m2 × v2i = (m1 + m2) × vf
Plug in the known values into the equation:
(2,000 kg) × (10.0 m/s) + (3,000 kg) × v2i = (2,000 kg + 3,000 kg) × 5.15 m/s
Simplifying the equation:
20,000 kg·m/s + 3,000 kg × v2i = 5,000 kg × 5.15 m/s
20,000 kg·m/s + 3,000 kg × v2i = 25,750 kg·m/s
Now, isolate the unknown variable.
3,000 kg × v2i = 25,750 kg·m/s - 20,000 kg·m/s
3,000 kg × v2i = 5,750 kg·m/s
Divide both sides of the equation by 3,000 kg:
v2i = 5,750 kg·m/s / 3,000 kg
v2i = 1.9167 m/s
Therefore, the speed of the 3,000-kg car before the collision is approximately 1.92 m/s.