Question

A 0.11 kg ball of dough is thrown straight up into the air with an initial speed of 17 m/s. The acceleration of gravity is 9.8 m/s 2 . What is its momentum at its maximum height? Answer in units of kg m/s. 002 (part 2 of 2) 1.0 points What is its momentum halfway to its maximum height? Answer in units of kg m/s.

Answer 1

Answer:

a) Its momentum at maximum height is 0 kg m/s

b) Its momentum at halfway maximum height is 1.32 kg m/s

Explanation:

The magnitude of momentum (p) is the product between mass (m) and velocity (v):

$p=mv$ (1)

a) At maximum height the velocity of the ball is instantly zero so:

$p=(0.11)(0)= 0\frac{kg\,m}{s}$

b) We can find the maximum height of the object with the kinematic equation:

$v^{2}=v_{0}^{2}+2ah$ (2)

with v the final speed, vo the initial speed, a the acceleration (a=-g=-9.8$\frac{m}{s^{2}}$) and h the height:

Using the fact that for $h_{max}$ v=0

$0=17^{2}-2(9.8)h_{max}$

Solving for $h_{max}$:

$h_{max}=\frac{-17^{2}}{2*9.8}=14.74m$

Now, halfway of maximum height is $\frac{h_{max}}{2}=7.37m$

We can use this value on (2) to find the speed at that point:

$v=\sqrt{v_{0}^{2}+2ah}=\sqrt{17^{2}-2(9.8*7.37)}$

$v=12.02\frac{m}{s}$

So, by (1) the moment is:

$p=(0.11)(12.02)=1.32\frac{kg\,m}{s}$