Find the initial speed of the bullet?

Question

A(n) 9 g bullet is fired into a 316 g block that is initially at rest at the edge of a frictionless table of height 1.8 m. The bullet remains in the block, and after impact the block lands
1.9 m from the bottom of the table. The acceleration of gravity is 9.8 m/s^2

Find the initial speed of the bullet. Answer in units of m/s.

Answer 1

You do the work.
(1) Calculate how long it takes for the block to fall 1.8 meters under the acceleration of gravity (remember that s=at²/2 so t = √(2*s/a) )
(2) Calculate the initial horizontal velocity of the block and bullet by dividing the distance it traveled (1.9 meters) by the time it took (which you calculated in (1))
(3) Now that you know the mass and velocity of the block and bullet, you can calculate the kinetic energy (Ek = mv²/2). (Remember: The mass is the block + the bullet)
(4) Since all of the energy in the system came from the bullet, you can now calculate the velocity of the bullet since it had the same kinetic energy as you calculated in (3) (v = √(2*Ek/m)) (Remember: In this calculation, the mass is that of the bullet *only*)

See how easy that is? ☺

Doug

Answer 2

The knowledge vigour of the pendulum and bullet on the maximum factor is two.78933 * nine.eight * zero.0635 Joules = one million.7358 J This have got to be same to the preliminary kinetic vigour of the bullet. KE = one million/two mv^two one million/two * zero.00933 v^two = one million.7358 v^two = 372.09 v = 19.2896 m/s