a man throws a ball of mass 0.5kg vertically upwards with a velocity of 25 m/s.find:[a}the initial momentum of the ball{b}momentum of the ball at the half way mark of the maximum height {given g=10 m/s square}
To find the initial momentum of the ball "a", we can use the formula:
Momentum (p) = mass (m) × velocity (v)
Given:
mass of the ball (m) = 0.5 kg
velocity of the ball (v) = 25 m/s
Substituting the given values into the equation, we get:
Momentum (p) = 0.5 kg × 25 m/s
Momentum (p) = 12.5 kg·m/s
Therefore, the initial momentum of the ball is 12.5 kg·m/s.
To find the momentum of the ball at the halfway mark of the maximum height "b", we can use the concept of conservation of momentum.
At the halfway mark, when the ball reaches its maximum height, its vertical velocity becomes zero. However, the momentum is conserved, so the momentum at the top is equal to the initial momentum.
Therefore, the momentum of the ball at the halfway mark of the maximum height is also 12.5 kg·m/s.
To find the initial momentum of the ball (a), we use the equation:
Momentum = Mass x Velocity
Given:
Mass (m) = 0.5 kg
Velocity (v) = 25 m/s
So, the initial momentum (p) of the ball is calculated as follows:
p = m x v
= 0.5 kg x 25 m/s
= 12.5 kg·m/s
Therefore, the initial momentum of the ball is 12.5 kg·m/s.
Now, to find the momentum of the ball at the halfway mark of the maximum height (b), we need to determine the maximum height that the ball reaches.
To do so, we can use the kinematic equation:
v^2 = u^2 + 2aS
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (25 m/s)
a = acceleration due to gravity (g = 10 m/s^2)
S = displacement or maximum height
We can rearrange the equation to solve for S:
S = (v^2 - u^2) / (2a)
S = (0^2 - 25^2) / (2 * -10)
= (0 - 625) / (-20)
= 625 / 20
= 31.25 m
Therefore, the maximum height the ball reaches is 31.25 meters.
Now, to find the momentum of the ball at the halfway mark, we can use the concept of conservation of momentum. The momentum of an object remains constant unless acted upon by an external force. Thus, the momentum at the halfway mark will be the same as the initial momentum.
Hence, the momentum of the ball at the halfway mark of the maximum height is also 12.5 kg·m/s.
initial momentum=.5*25 kg m/s
at halfway up, it has lost 1/2 of its energy.
energy then=initial energy.
1/2 m v^2=1/2 *1/2 m 25^2
so v= .7*25
momentum then= .7 of initial momentum