Physic homework (kinematics, Newton's Laws and Relative motion/velocity)

[NiteCrawler]

I have this homework problem that I really need help on. Thank you for your time

Train A is traveling down a long straight track at a speed of 30m/s. The train A is 0.40km long and the last car at the very end of the train has a cannon mounted on it. Neglect the height of the cannon with respect to the ground. Unbelievably, the cannon aimed at an angel of 50dregrees above the horizontal and in the direction of the train’s motion. The cannon fires a ball at a speed of 55m/d relative to the train. The firing of the cannon does not affect the train speed; i.e., neglect recoil. The conductor of train A has to make a quick decision.

a) Assuming no air resistance, at what time will the ball land?
b) Where does the ball hit the track relative to where it was launched? If the conductor does not change the train’s speed, where will the front and back of the train be at the time of the ball lands?
c) If the conductor chooses to put on brakes at time t=0, what acceleration is required so that the train just stops before meeting the crater left by the ball hitting the track?
d) Draw a graph of horizontal position versus time for the ball, the front of the train, and the back of the train for the case that the train continues with constant velocity and for the case that it just stops before it reaches the crater. On our graph, indicate the final and initial positions of the ball, the front of the train, and the back of the train

 

[Galileo]

a) Assuming no air resistance, at what time will the ball land?

You might as well view this problem in the train's frame of reference.
The vertical component of the initial velocity is 55sin(50) m/s. Now use the 'freefall formula' $y=v_0t-\frac{1}{2}gt^2$

b) Where does the ball hit the track relative to where it was launched? If the conductor does not change the train’s speed, where will the front and back of the train be at the time of the ball lands?

The speed of the ball relative to the ground is the speed of the ball relative to the train plus the speed of the train relative to the track. You know the flight time, so you can calculate the distance.

c) If the conductor chooses to put on brakes at time t=0, what acceleration is required so that the train just stops before meeting the crater left by the ball hitting the track?

This is the acceleration required to come to a halt after the distance the ball has traveled. You got that distance in b)

d) Draw a graph of horizontal position versus time for the ball, the front of the train, and the back of the train for the case that the train continues with constant velocity and for the case that it just stops before it reaches the crater. On our graph, indicate the final and initial positions of the ball, the front of the train, and the back of the train

Why are you asking us how to draw this graph? Especially after you know the positions of the ball and train after doing the previous exercises.

 

[wais]

Hi there,

Thank you for reaching out for help with your physics homework. I would be happy to assist you with this problem.

To begin, let's first define some variables:
- Train A's speed: v = 30m/s
- Train A's length: L = 0.40km = 400m
- Angle of cannon: θ = 50°
- Ball's initial speed relative to train: u = 55m/s
- Time: t
- Acceleration: a

a) To find the time at which the ball will land, we can use the kinematic equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this case, the displacement of the ball is the length of the train, 400m. The initial velocity is 55m/s, and we can assume that the acceleration is due to gravity, which is -9.8m/s^2. So, we have 400 = 55t + (1/2)(-9.8)t^2. Solving for t, we get t = 13.6 seconds. Therefore, the ball will land after 13.6 seconds.

b) The ball will hit the track at a horizontal distance of 13.6 x 55 = 748m from the initial position of the cannon. If the conductor does not change the train's speed, the front and back of the train will still be at their initial positions when the ball lands. However, if the conductor chooses to put on brakes at t=0, the train will experience a deceleration in order to stop before reaching the crater. To calculate the required acceleration, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, v = 0 (since the train needs to stop), u = 30m/s, and s = 400m. Solving for a, we get a = -0.225m/s^2. Therefore, the train needs to decelerate at a rate of 0.225m/s^2 in order to stop before reaching the crater.

c) The graph for the ball's horizontal position versus time will