Question

These problems involve Impulse-Mometum theorem, and the Work-Kinetic Energy theorem. Both theorems are combinations of Newton's Second Law and 1D kinmatics. DON"T use scientific notation in all answers. (a) A runaway train car that has a mass of 16000 kg travels at a speed of 9.8 m/s down a track. Compute the time (in seconds) required for a force of 1900 N to bring the car to rest. 0 s Incorrect (0.0%) (4 attempts remaining)Input 1 StatusIncorrect (0.0%)4 attempts remaining. (b) A bullet is accelerated (from rest) down the barrel of a gun by hot gases produced in the combustion of gun powder. If the average force exerted on a 0.027 kg bullet to accelerate it is 3500 N, what will be the bullet's speed (in m/s) in a time of 0.004 s?

Answer 1

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:

$F\Delta t = m\Delta v$

where

F is the force applied

$\Delta t$ is the time interval

m is the mass of the object

$\Delta v$ 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)

$\Delta v = 0 -9.8 m/s = -9.8 m/s$ is the change in velocity of the car

Solving for $\Delta t$, we find the time needed:

$\Delta t = \frac{m\Delta v}{F}=\frac{(16000)(-9.8)}{-1900}=82.5 s$

b)

Again, in this part we can also use the impulse-momentum theorem:

$F\Delta t = m\Delta v$

where

F is the force applied

$\Delta t$ is the time interval

m is the mass of the object

$\Delta v$ 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

$\Delta t = 0.004 s$ is the time interval

Solving for $\Delta v$, we find the change in velocity of the bullet:

$\Delta v = \frac{F \Delta t}{m}=\frac{(3500)(0.027)}{0.004}=23,625 m/s$

And since the initial velocity of the bullet is zero, the final velocity (and speed) is

$v=23,625 m/s$

Learn more about impulse and momentum:

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