In a ballistics test, a 1.50-g bullet is fired through a 28.0-kg block traveling horizontally toward the bullet. In this test, the bullet takes 11.4 ms to pass through the block as it reverses the block's velocity from 1.75 m/s to the right to 1.20 m/s to the left with constant acceleration. Find the magnitude of the force that the bullet exerts on the block during this ballistics test.

Respuesta :

By newtons 3rd law of motion

Rate of change in momentum of the bullet = Rate of change in momentum of the block

ForceBullet=Forceblock
Force∙=mblock∗(vfinal−vinitial)/t
ForceBullet=28(1.20−(−1.75)/11.4×10−3