One man hold a 3kg iron ball fall from 1.5m in the air. The ball then fall on to a table .The table is 0.5m tall. The ball bound up 3cm up and fall again. What is the impact(in Newton) the ball hit the table on the first time?
H=1.5-0.5 = 1 m, h=0.03 m
m•g•H=m•v1²/2
v1=sqrt(2gH) = sqrt(2•9.8•1)= 4.43 m/s.
m•v2²/2=mgh.
v2= sqrt(2gh) = sqrt(2•9.8•0.03)= 0.77 m/s.
Δp=p2-p1=mv2 –(-mv1) = m(v2+v1) =3•(4.43+0.77) =15.6 kg•m/s.
If you knew time of impact “Δt”,
F•Δt= Δp,
F= Δp/ Δt (in Newtons)
A ball is held at the top of a table. The person holding the ball drops it, and the ball is allowed to fall toward Earth. Answer the following questions about the ball.
a. When the ball is held at the top of the table (before being dropped), what type of energy does the ball have?
(1 point)
Responses
motion energy
motion energy
electrical energy
electrical energy
kinetic energy
kinetic energy
potential energy
To calculate the impact force that the ball exerts on the table when it hits for the first time, we need to consider the change in momentum of the ball.
The change in momentum is given by the equation:
Δp = m * Δv
Where:
Δp = Change in momentum
m = Mass of the ball
Δv = Change in velocity of the ball
We can break down the problem into two parts: the falling motion and the rebound motion.
First, let's calculate the change in velocity when the ball falls from a height of 1.5m to a height of 0.5m.
Using the equation for gravitational potential energy:
m * g * Δh = 0.5 * m * Δv^2
Where:
m = Mass of the ball (3kg)
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Δh = Change in height (1.5m - 0.5m = 1m)
Solving for Δv:
Δv = √(2 * g * Δh)
Δv = √(2 * 9.8 * 1) = √(19.6) = 4.43 m/s
Now, let's calculate the change in momentum:
Δp = m * Δv = 3kg * 4.43 m/s = 13.29 kg·m/s
The impact force is equal to the rate of change of momentum, so we'll divide the change in momentum by the time it takes for the collision to occur.
Since the table rebounded the ball 3cm (0.03m), we have to consider that the ball traveled upward and downward, so the total distance traveled is 0.06m.
To find the time it takes for the ball to cover this distance, we can use the equation:
Δh = 0.5 * g * t^2
Where:
Δh = Change in height (0.06m)
g = Acceleration due to gravity (approximately 9.8 m/s^2)
t = Time
Solving for t:
0.06 = 0.5 * 9.8 * t^2
t^2 = 0.06 / (0.5 * 9.8)
t^2 = 0.00612
t ≈ √0.00612
t ≈ 0.078s
Finally, we can calculate the impact force using the formula:
F = Δp / t
F = 13.29 kg·m/s / 0.078s
F ≈ 170 N
Therefore, the impact force that the ball hits the table for the first time is approximately 170 Newtons.