A car P is moving with uniform speed of 5m/s towards a carriage of mass 9kg at rest kept on the rails at point B. the height of AC is 120m.cannon balls of 1kg are fired from the car with an initial velocity of 100m/s at an angle 30 degree with the horizontal.the first cannon ball hits the stationary carriage after a time t' and stick to it. determine t'. the second cannon ball is fired . assume that resistive forces between the rails and the carriage is constant and ignore the vertical motion of carriage . if the second ball also hits and sticks to the carriage. what will be the horizontal velocity of carriage just after the second impact. DIAGRAM:

|--|
car--> 0--0 C
--------|
| ( )<--carriage
| 0-0
A|------------B

To determine the time t' when the first cannonball hits the stationary carriage, we can use the following steps:

1. Calculate the horizontal component of the initial velocity of the cannonball fired from the car using the given angle of 30 degrees. We can use the equation:

v0x = v0 * cos(angle)

where v0 is the initial velocity of the cannonball (100 m/s) and angle is the given angle (30 degrees).
v0x = 100 m/s * cos(30°) = 100 m/s * √(3)/2 ≈ 86.6025 m/s

2. Now, we need to find the time it takes for the cannonball to travel the horizontal distance BC. We can use the equation:

distance = speed * time

The distance BC is equal to the height of AC, which is 120 m. The speed is the horizontal component of the initial velocity of the cannonball, v0x. Let's denote the time as t.

120 m = 86.6025 m/s * t

Solving this equation for t gives us:

t = 120 m / 86.6025 m/s ≈ 1.3841 s

Therefore, t' is approximately 1.3841 seconds.

To determine the horizontal velocity of the carriage just after the second impact, we can use the principle of conservation of momentum. Since the cannonball sticks to the carriage, the total momentum before and after the collision should be the same.

1. Calculate the momentum of the car before the collision:

momentum_car = mass_car * velocity_car

Since the car is moving with a uniform speed of 5 m/s, the velocity_car is 5 m/s.

momentum_car = mass_car * 5 m/s

2. Calculate the momentum of the second cannonball before the collision:

momentum_cannonball = mass_cannonball * velocity_cannonball

The mass of the cannonball is given as 1 kg, and the velocity_cannonball is the horizontal component of the initial velocity (v0x) of the cannonball.

momentum_cannonball = 1 kg * 86.6025 m/s ≈ 86.6025 kg⋅m/s

3. Add up the momenta of the car and the cannonball before the collision:

total_momentum_before = momentum_car + momentum_cannonball

4. After the collision, since the cannonball sticks to the carriage, the momentum of the combination (car + carriage + cannonball) is equal to the momentum of the first cannonball.

momentum_combination = mass_cannonball * velocity_carriage

Solving this equation for the velocity_carriage gives us:

velocity_carriage = total_momentum_before / mass_cannonball

Plugging in the values, we get:

velocity_carriage = (momentum_car + momentum_cannonball) / mass_cannonball

velocity_carriage = (mass_car * 5 m/s + 1 kg * 86.6025 m/s) / 1 kg

Therefore, the horizontal velocity of the carriage just after the second impact is the calculated value from this equation. Substitute the given values to obtain the final answer.