a) The time needed to stop the car is 82.5 s
b) The final speed of the bullet is 23,625 m/s
Explanation:
a)
We can solve this part of the problem by using the impulse-momentum theorem, which states that:
"The impulse exerted on an object (the product between force applied and time interval) is equal to the change in momentum of the object"
Mathematically:
[tex]F\Delta t = m\Delta v[/tex]
where
F is the force applied
[tex]\Delta t[/tex] is the time interval
m is the mass of the object
[tex]\Delta v[/tex] is the change in velocity
For the train car in this problem, we have
m = 16000 kg is the mass
F = -1900 N is the force applied (with negative sign, since it is applied in the direction opposite to the direction of motion, in order to stop the train)
[tex]\Delta v = 0 -9.8 m/s = -9.8 m/s[/tex] is the change in velocity of the car
Solving for [tex]\Delta t[/tex], we find the time needed:
[tex]\Delta t = \frac{m\Delta v}{F}=\frac{(16000)(-9.8)}{-1900}=82.5 s[/tex]
b)
Again, in this part we can also use the impulse-momentum theorem:
[tex]F\Delta t = m\Delta v[/tex]
where
F is the force applied
[tex]\Delta t[/tex] is the time interval
m is the mass of the object
[tex]\Delta v[/tex] is the change in velocity
For the bullet in this problem, we have:
m = 0.027 kg is the mass
F = 3500 N is the force applied
[tex]\Delta t = 0.004 s[/tex] is the time interval
Solving for [tex]\Delta v[/tex], we find the change in velocity of the bullet:
[tex]\Delta v = \frac{F \Delta t}{m}=\frac{(3500)(0.027)}{0.004}=23,625 m/s[/tex]
And since the initial velocity of the bullet is zero, the final velocity (and speed) is
[tex]v=23,625 m/s[/tex]
Learn more about impulse and momentum:
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