Phyrics hw help.?

Question

A 0.007-kg bullet traveling horizontally at 400.4 m/s strikes a 4.1-kg block of wood sitting at the edge of a table. The bullet is lodged into the wood. If the table height is 1.2 m, how far from the table does the block hit the floor?

I thought i had to use v=sqrt (2gh)... but clearly i that is not the right equation to use. any help would be great! thanks!

Answer 1

First you follow the action, then you decide what equation to use. The first event, the bullet hitting the block of wood moves the block. Assuming momentum is conserved:
m1V1 of the bullet = m2V2 of the block
We know m1 and V1, and that m2>>m1, so we can find V2 of the block. This causes the block to move in a horizontal direction at V2 speed until it drops to the floor, a distance of 1.2 m. The time for this drop is independent of the horizontal velocity, so 1.2 = 1/2 gt^2. I believe g=9.8 m/sec^2, so we can solve for t, which is about 1/2 second. So the distance from the table is V2 x t.

Answer 2

use conservation of energy laws and projectile motion.

in this case it would be mass of bullet*velocity = (mass of bullet + mass of block)*your unknown velocity

just solve for the unknown velocity

now you must break that velocity into components using these equations

y = vt + 1/2at ^2
x = vt+ 1/2at^2

your initial velocity in the y equation is 0 so vt cancels out. Your acceleration is that of gravity (9.8) and you can now get your time by plugging into y the hieght of the table

1.2=0+1/2(9.8)t^2

solve for t

now take that t value and plug it into your x equation

remember that velocity from the momentum equation?
use that for the v in the vt term and also input the t from the y equation in that t. and the value of x will be your final answer.