A person holding a 15.0 kg containing one 50.0 g bullet is riding on a train that is traveling at 75.0 km/h east. If the man fir
1 answer:
Answer:
The velocity of the gun relative to the ground is 19.66 m/s
Explanation:
Given data,
The mass of the gun, M = 15.0 kg
The mass of the bullet, m = 50 g
The velocity of the train, v = 75 km/h
= 20.83 m/s
The velocity of bullet relative to train, V' = 350 m/s
The velocity of bullet relative to ground, V = 350 + 20
= 370 m/s
According to the law of conservation of momentum,
Mv' + mV' = 0
= -1.17 m/s
Therefore, the velocity of the gun with,
v₀ = V + v'
= 20.83 - 1.17
= 19.66 m/s
Hence, the velocity of the gun relative to the ground is 19.66 m/s
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Answer:
a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.
b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.
c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.
Please, see the attached figures and the explanation for a description of the figures.
Explanation:
Hi there!
The equations for the height and velocity of the stone are as follows:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height
y0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity at time t
a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:
y = y0 + v0 · t + 1/2 · g · t²
11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 18 m/s · t - 11 m
Solving the quadratic equation:
t = 0.7743 s and t = 2.899 s
(Notice that I have used more significant figures to avoid error by rounding)
The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of 11 m for the first time.
b) Let´s use the equation of velocity:
v = v0 + g · t
at t = 0.77443 s
v = 18 m/s - 9.8 m/s² · 0.77443 s
v = 10.41 m/s
at t = 2.899 s
v = 18 m/s - 9.8 m/s² · 2.899 s
v = - 10.41 m/s
(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)
c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.
Figures
The functions to plot are the following:
height in function of time (figure 1, x-axis: time. y-axis: height)
y = -4.9t² + 18t
velocity in function of time (figure 2, x-axis: time. y-axis velocity)
v = -9.8t + 18
Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)
a = -9.8
