Question

7. A 1.00 kg rock is thrown up into the air from ground level at a speed of 8.00 m/s. The ball travels up
to a maximum height, then returns to the ground. Calculate the rock’s momentum as it strikes the
ground.

Answer 1

The rock was thrown from ground level, and it will return to ground level. Using Conservation of Energy, the amount of kinetic energy it has when it's initially thrown upward will be equal to the amount of kinetic energy it has just before it hits the ground. This tells us that the velocity of the rock has the same magnitude as when it was initially thrown.
\[\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2\]
\[v_i^2=v_f^2 \rightarrow |v_i|=|v_f|\]
Since the rock will be falling straight back downwards (assuming it was thrown straight upwards initially), we see that:
\[v_f=-v_i\]
Therefore, the rocks momentum is given by:
\[p = mv_f = -mv_i \]

Answer 2

i thought that this question would have something to do with projectile motion.....

Answer 3

b

Answer 4

Well, if we relax the condition that the ball is thrown straight into the air, you'll actually arrive at the same answer.

The velocity in the x direction doesn't change, as there is no acceleration. The velocity in the y direction will have the same magnitude, but opposite sign, when the ball comes down.

Another way to look at it is through conservation of energy. Initially, we define ground level to have zero potential energy, so all the energy is kinetic. As the ball rises, some of this kinetic energy gets converted to potential energy, and then the ball falls back to toward the ground. When the ball reaches ground level, it will have all it's energy as kinetic energy once again.

\[Energy = \frac{p^2}{2m}, E_i = E_f\]\[p_i^2 = p_f^2\]\[|p_i| = |p_f|\]