Answer:
The change in momentum = -20000 kg m/s.
Explanation:
Mass m = 1000 kg
speed v₁ = 20 m/s
speed v₂ = 0 m/s
We know that,
The change in momentum
ΔP = m (Δv)
ΔP = m (v₂ - v₁)
= 1000 (0 - 20)
= 1000 (-20)
= -20000 kg m/s
Thus, the change in momentum = -20000 kg m/s.
Note: negative sign indicates that the velocity is reducing when it hits the barrier.
Answer:
The ball will have a kinetic energy of 0.615 Joules.
Explanation:
Use the kinetic energy formula
The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)
Answer:
- <u>77.8 m/s, downward</u>
Explanation:
For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf
- Average speed = (Vf + Vi)/2
Also, by definition, the average speed is the distance divided by the time:
- Average speed = distance / time
Then:
Other kinematic equation for uniform acceleration is:
Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)
Replacing the known values we can set a system of two equations:
From (Vf + Vi)/2 = 300m/6.62s
(Vf + Vi) = 2 × 300m/6.62s
- Vf + Vi = 90.634 equation 1
From Vf = Vi + a×t
Vf - Vi = 9.8 (6.62)
- Vf - Vi = 64.876 equation 2
Adding the two equations:
- Vf = 77.8 m/s downward (velocities must be reported with their directions)
