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.

Question

anonymous · Answer

By newtons 3rd law of motion
Rate of change in momentum of the bullet = Rate of change in momentum of the block
$$Force_{Bullet} = Force_{block}$$
$$Force_{\bullet} = \frac{m_{block} *(v_{final}-v_{initial})}{t}$$
$$Force_{Bullet} = \frac{28(1.20-(-1.75)}{11.4 \times 10^-3}$$

anthonyn2121 · Answer

Is this question possible to do without using any momentum equations? My class has not covered this section yet, and I think we are only supposed to be using forces.

vincent-lyon.fr · Answer

There is no way solving this problem not using the change in momentum of the blocks.

btw, "a 1.50-g bullet" should not be hyphenated (as stated in official SI brochure)