Question

A 32 g bullet is fired horizontally from a rifle at 380 m/s into a water bottle at y pointblank range. The bullet travels completely through the 12 cm long bottle, located 2 m above the ground. (You may assume there is no change in the height of the bullet while travelling through the bottle.) It finally hits the ground at a point 3 m downrange from the point where it left the water. Determine: the average force exerted by the w after on the bullet. the total time elapsed from when the rifle is fired to when the bullet hits the ground.

Answer 1

a )

F is the average force excerted by water on bullet

F X d X cos180 = 0.5 X m ( V₂² - V₁² )

F X 0.12 X ( -1 ) = 0.5 X 0.032 (0² - 380² )

F X ( -0.12 ) = -2310.4

F = 19253.33 N

b )

the acceleration of the bullet a = - F / m

= - 19253.33 / 0.032

a = - 601666.66 m/s²

the time is = ( V₂ - V₁ ) / a

= ( 0 - 380 ) / ( - 601666.66 )

= 0.631 ms

and t = 2 h / g

t = ( 2 X 2 / 9. 8

t = 0.6388 sec

the total time is t = 0.631 X 10^-3 + 0.6388

T = 0.6394 sec