Question

A skater boy is standing in front of a wall facing towards it. He throws the ball horizontally to the wall with an initial velocity of 8 m/s relative to the wall. The ball starts its motion at a height of 1.8 m and its mass is 4 kg. If the distance between the wall and the boy is 3.2 m and boy's mass is 32(collision is elastic)

a) Calculate the angle between the ball and the wall when it hits the wall

Answer 1

@douglaswinslowcooper sir could you take a look at this when you are available, i keep on finding bad results :(

Answer 2

1. find horizontal and vertical speed of the ball when it touches the wall
2. angle = inverse tan (vertical speed/horizontal speed)

Answer 3

the horizontal speed is constant so i only need to find the vertical one right? Also we find the time that the ball hits the ball as 0.4 sec? Then i tried to use y=y(initial)+(V(initial)*t)+((1/2)a(y)t^2)) with t=0.4 V(initial) as 0 since it has no vertical velocity initially and a as 9.81. But somehow i found nothing helpful

Answer 4

came to my mind that for y direction i can use v=a*t right?

Answer 5

Yes, horizontal component of velocity is constant, so that gives you the time t to reach the wall. The vertical component will be v = - g t = - (9.8 m/s^2) t

The tangent of the angle of the ball (horizontal =0 o) is tan(angle)= -v t / 8.
That gives you the angle with horizontal, which you can adjust to give angle with vertical.

Answer 6

okay i found the angle arctan(2) now and it seems reasonable. Thanks A lot!

Answer 7

additionally question asks " Calculate the momentum transfer to the wall (in kg m/s ) but i dont understand doesn't the ball gives all of his momentum to the wall?

Answer 8

On elastic rebound, twice the momentum, 2 mass x velocity in x direction is exchanged, If sticks to the wall, inelastic collision, then just one mass x velocity in x direction is given up. Takes more force to reverse direction than just to stop ball.