A small metal ball with a mass of m = 78.9 g is attached to a string of length l = 1.76 m. It is held at an angle of = 44.5° with respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 2.72 m, and then later it hits the ground. At what distance x will the ball land?
draw the figure:
initial PEnergy=mgl(1-cosTheta)
final KE=mgl(1-cosTheta)=1/2 mv^2
solve for v.
Then, that is also the hit ball velocity (conservation of momentum).
time to fall h,
h=1/2 g t^2
t=sqrt(2h/g)
distance=velocity*time
To find the distance at which the ball will land, we need to split the problem into two parts: the vertical motion and the horizontal motion.
First, let's consider the vertical motion. The ball is initially released at a height h = 2.72 m above the ground, so the initial vertical velocity is zero. The only force acting on the ball in the vertical direction is gravity, which causes the ball to accelerate downward.
Using the kinematic equation for vertical motion:
h = (1/2) * g * t^2
where h is the initial height (2.72 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the ball to hit the ground.
Solving for t, we get:
t = √(2h/g)
= √(2 * 2.72 m / 9.8 m/s^2)
≈ 0.745 s
Now that we know the time it takes for the ball to hit the ground, let's consider the horizontal motion. The second ball flies off with a horizontal initial velocity. Since the collision between the two balls is perfectly elastic, the initial horizontal velocity of the second ball must be the same as the final horizontal velocity of the first ball.
We can find the initial horizontal velocity of the second ball using trigonometry. The angle at which the first ball is released is given as θ = 44.5°. Since the initial vertical velocity is zero, the initial horizontal velocity (v0x) is equal to the initial velocity of the first ball multiplied by the cosine of the angle.
v0x = v * cos(θ)
The initial velocity of the first ball can be calculated using its height h and the time it takes to reach that height. We can use the equation:
h = v0y * t + (1/2) * g * t^2
Since the initial vertical velocity v0y is zero, this simplifies to:
h = (1/2) * g * t^2
Solving for v0y, we get:
v0y = (2h/g)^0.5
≈ 2.951 m/s
Now we can calculate the initial horizontal velocity:
v0x = (v0y) * cos(θ)
≈ (2.951 m/s) * cos(44.5°)
≈ 2.097 m/s
Knowing the initial horizontal velocity, we can use the equation of motion for horizontal motion:
x = v0x * t
Plugging in the values:
x ≈ (2.097 m/s) * (0.745 s)
≈ 1.561 m
Therefore, the ball will land at a distance of approximately 1.561 meters.