A rifle shoots a 4.90 g bullet out of its barrel. The bullet has a muzzle veloci

Question

A rifle shoots a 4.90 g bullet out of its barrel. The bullet has a muzzle velocity of 970 m/s just as it leaves the barrel.

A)Assuming a constant horizontal acceleration over a distance of 41.0 cm starting from rest, with no friction between the bullet and the barrel, what force does the rifle exert on the bullet while it is in the barrel?

Answer is: F = 5620 N

B) Draw a free-body diagram of the bullet while it is in the barrel.

C) Draw a free-body diagram of the bullet justafterit has left the barrel.

a = ______ g

E) For how long a time is the bullet in the barrel?

t = _____ s

Explanation / Answer

v^2 = u^2 + 2as
970^2 = 0^2 + 2*a*0.41

a = 1147439.024390244 m/s^2

a = 1147439.024390244m/s^2 (to three significant figures)

a = 116966.2614057333 g

You asked for a force also, and that can be found as follows:

F = ma
F = 0.0049 * 1147439

F = 5616.719872703314N (to three significant figures)

And also the time in the barrel:

s = t(u + v)/2

0.49 = t * 970/2

t = 0.0010103092783505s

t = 0.0010103092783505 s

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