1.A truck with a mass 5000 kg is heading East at 20 m/s. Meanwhile a car with a mass 2500 kg is heading North at 30 m/s. The two vehicles collide at an intersection, get stuck together, and the wreckage keeps moving to some direction.find the speed of the wreckage and the direction it moves to, just after the collision.

2.A horizontal disk (M = 150 g, R = 20 cm, I = MR*2/2 )rotates at 10 rev/min about its axis.(no friction, no torque from outside). An overweight flying bug (m = 15 g) lands vertically on the disk at a point half way to the center. Use the conservation of angular momentum to find the angular speed of the plate just after this collision.

1. Use the fact that the total momentum of truck and car are the same before and after the collision. You will need to consider momenta in both East direction (x axis) and North direction (y axis). We will be happy to critique your work.

2. Use the fact that the toal angular momentum about the disc axis, before and after the bug lands, will be the same.

I w1 = I w2 + m (R/2)^2 w2

(w2/w1) = I/[I + m R^2/4]
= 1/[1 + (m/I) R^2/4]
= 1/[1 + (m/M)(R^2/4)/(R^2/2)]
= 1/[1 + (1/2)(m/M)]

w is the angular speed. w2/w1 is the ratio by which it decreases.

1. Truck moving along east along X direction and car moving north along the Y direction.

Total momentum along the X direction:
5000 * 20 + 0 = 7500 V1 cosine ө
1000 = 75 V1 cosine ө --------------------- (1)
Total momentum along the Y direction:
2500 * 30 + 0 = 7500 V1 sin ө
750 =75 V1sin ө------------------------- (2)
Dividing (2) and (1),
Tan ө=0.75
Ө =370
From equation (2),
V1 sin ө = 10
V1 sin 370 = 10
V1 = 10/sin 370
V1 = 10/0.6
V1 = 16.6 m/s is the velocity of the combined vehicles

To solve these problems, we can use the principle of conservation of momentum and conservation of angular momentum, respectively. Here's how we can approach each problem:

1. For the collision of the truck and the car, we can analyze the momentum vectors for each vehicle before and after the collision. Since momentum is a vector quantity, we need to consider both magnitude and direction.

Before the collision:
- Momentum of the truck: P1 = m1 * v1 = (5000 kg) * (20 m/s) = 100,000 kg m/s (heading East)
- Momentum of the car: P2 = m2 * v2 = (2500 kg) * (30 m/s) = 75,000 kg m/s (heading North)

After the collision, the two vehicles stick together and move as one:
- Let's assume the resulting velocity of the wreckage is V, and the angle it makes with the positive x-axis is θ.

Using conservation of momentum, we can equate the initial total momentum to the final total momentum:
P1 + P2 = (m1 + m2) * V

Breaking this equation into components:
P1x + P2x = (m1 + m2) * V * cos(θ) -- for the x-component
P1y + P2y = (m1 + m2) * V * sin(θ) -- for the y-component

From the given information, we know P1x = 100,000 kg m/s (since it's heading East) and P2y = 75,000 kg m/s (since it's heading North). P1y and P2x are both zero since there is no velocity in those directions initially.

Solving these equations, we can find V (the speed of the wreckage) and θ (the direction it moves to) after the collision.

2. For the collision between the disk and the flying bug, we can analyze the angular momentum before and after the collision.

Before the collision:
- Angular momentum of the disk: L1 = I * ω1 = (MR²/2) * (2πn/60) -- where n is the rotational speed in revolutions per minute (rpm)

After the collision, the bug lands vertically on the disk:
- Let's assume the angular speed of the disk just after the collision is ω2.

Using the conservation of angular momentum, we can equate the initial angular momentum to the final angular momentum:
L1 = L2
(I * ω1) = [(I + mr²) * ω2] -- since the bug lands halfway to the center, its distance from the axis of rotation is r

From the given information, we know I = (MR²/2), m = 15 g = 0.015 kg, R = 0.20 m, and ω1 = 10 rev/min = (10 * 2π/60) rad/s.

Solving this equation, we can find ω2 (the angular speed of the disk just after the collision).

By following these steps and plugging in the given values, you should be able to calculate the required speeds and directions.