A 0.19 kg ball of dough is thrown straight up into the air with an initial speed of 13 m/s.The acceleration of gravity is 9.81 m/s.
b) What is its momentum halfway to its maximum height on the way up?
Answer in units of kg · m/s.
max height:(velocity is zero)
vf^2=vi^2+2ad
0=13^2-2g
max height=13^2/2g
halfway=13^2/4g
vf^2=vi^2+2ad'
vf^2=13^2 -2g*13^2/4g
v half way= sqrt(13^2/2)
momentum half way=mass times vhalfway.
To find the momentum halfway to its maximum height, we need to determine the velocity of the ball at that point.
The velocity of the ball at the halfway point can be calculated using the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (which is 0 since the ball reaches its maximum height)
u = initial velocity (given as 13 m/s)
a = acceleration (which is the acceleration due to gravity, -9.81 m/s²)
s = displacement (halfway to maximum height)
Rearranging the equation, we have:
0 = 13² + 2(-9.81)s
Simplifying further:
169 = -19.62s
Divide both sides by -19.62:
s = -8.62 m
Since the displacement cannot be negative in this context, we can ignore the negative sign and take the magnitude of the value, giving us s = 8.62 m.
Now, we can calculate the velocity at the halfway point:
v = u + at
v = 13 + (-9.81)(8.62)
v ≈ -17.76 m/s
Again, we take the magnitude of the velocity, ignoring the negative sign:
v ≈ 17.76 m/s
Finally, we can find the momentum at that point using the formula:
momentum = mass × velocity
Given:
mass (m) = 0.19 kg
velocity (v) = 17.76 m/s
momentum = 0.19 kg × 17.76 m/s
Calculating the product:
momentum ≈ 3.37 kg · m/s
Therefore, the momentum halfway to its maximum height on the way up is approximately 3.37 kg · m/s.