Collisions and Momentum: Solving for Initial Velocity

AI Thread Summary
The discussion focuses on solving for the initial velocity of a bullet in various collision scenarios with a block on a frictionless surface. For part A, the conservation of momentum is applied, leading to a formula for the initial velocity based on the final velocity of the block and the time taken to fall. In part B, a similar approach is used, adjusting for the bullet bouncing back at half its speed, resulting in a different expression for initial velocity. The participant is still working on part C, which involves the bullet passing through the block, and is seeking clarification on the calculations and concepts. The final comparison in part D aims to determine which scenario results in the block moving farther, the sticky or bouncy bullet.
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Homework Statement


A bullet of mass m moving with velocity V_o crashes into a block of mass M at rest on a frictionless horizontal surface the block is sitting at the edge of a table that is H high. When the bullet hits the block it flies off the table and it lands a distance x from the bottom of the table. In each case below, determine the initial velocity of the bullet in the terms of the give quantities.

A) in the first case assume that the bullet sticks to the block
B) in the second case assume that the bullet bounces back directly with half its original speed
C)In the Third case the bullet crashes through the block emerging on the other side with half its original speed but still moving in the same direction
D) which makes the block move farther, the sticky bullet or the bouncy bullet?
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Homework Equations


Was thinking i can use the conservation of momentum for part A

M_total*V=M_1*V_1+M_2*V_2

The Attempt at a Solution



A) So first i solve for the Final velocity
V=(m*v_o)/(M+m)

Now i know X = t*V (so i must solve for the time it takes for the box to hit the ground to get t)

h=(.5)at^2
a=g
So
t= (h/2g)^(1/2) (don't know how to make the square root symbol)

Now i just plug this in and get

X= ((m*v_o)/(M+m))*(h/2g)^(1/2)

Then solve for V_o

So V_0= (X(M+m))/(m*(h/2g)^(1/2))B) so i did the same thing i did for part A and used the same t

m(V_0)=m(.5V_0)+M(V)
V=(3m(v_0))/(2M)

X=t*(3m(v_0))/(2M)
t=(h/2g)^(1/2)

V_0= (2M*X)/(3m*(h/2g)^(1/2))
C)And i still haven't had time to start on Part C
D)For this i just got to compare the x= answers i got from part A and b and see what one is bigger (not sure how to do this either)
 
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Just curious if anyone could tell me if i have the right idea -Thanks Tom
 
A

h=(.5)at^2
a=g
So
t= (2h/g)^(1/2)
 
so are u saying that is right?
 
One thing I noticed for part b, is that you left a sign off, the bullet recoils and so its contribution is -1/2mv_0. But keep those results I think they are useful for c.
 
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