An object is moring in a negative direction with a negative acceleration, for example an object dropped out a vindow. What does
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Answer:
New Momentum of Ball B
<u>Explanation:</u>
Given:
Mass of Ball A=1kg
Mass of Ball B= 2kg
Mass of Ball C=5kg
Mass of Ball D=7kg
Velocities of A=B=C=D=2.2
Momentum of Ball A=2.2
Momentum of Ball B=4.4
Momentum of Ball C=11
Momentum of Ball D=15
To Find:
Change in Momentum When of Ball B gets tripled
Solution:
Though all balls have same velocity, thus we get
Velocities of A=B=C=D=2.2
Initial Momentum of Ball B=4.4
If the Mass of Ball B gets tripled;
We get New Mass of Ball B=3×Actual Mass of the ball
=3×2=6kg
Thus we get Mass of Ball B=6kg
According to the formula,
Change in momentum of Ball B
Where =change in momentum
m=mass of the ball B
=change in velocity ball B
And since all balls, have same velocity
Thus the above equation, changes to
Substitute all the values in the above equation we get
Result:
Thus the New Momentum of ball B
Answer:
Your friend has to wait 0.26 s after you throw the ball to start running.
Explanation:
The equation that gives the position vector of the ball is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal positon
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity
The equation of displacement of your friend is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of your friend at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.
Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:
y = y0 + v0 · t ·sin α + 1/2 · g · t²
-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m
Solving the quadratic equation:
t = 1.71 s
Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):
x= x0 + v0 · t · cos α
x = 0m + 9.00 m/s · 1.71 s · cos 33°
x = 12.9 m
Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.
Let´s see, how much time it takes your friend to run that distance:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)
x = 1/2 · a · t²
1.9 m = 1/2 · 1.80 m/s² · t²
Solving for t
t = 1.45 s
Then, since the time of flight of the ball is 1.71 s, your friend has to wait
1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.
