Collisions and Momentum: Solving for Initial Velocity

In summary, the bullet in part C crashes through the block and emerges on the other side with half its original speed.
  • #1
tomrule007
18
0

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?
physics.jpg

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)
 
Last edited:
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  • #2
Just curious if anyone could tell me if i have the right idea -Thanks Tom
 
  • #3
A

h=(.5)at^2
a=g
So
t= (2h/g)^(1/2)
 
  • #4
so are u saying that is right?
 
  • #5
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.
 

Related to Collisions and Momentum: Solving for Initial Velocity

What is the definition of a collision?

A collision occurs when two or more objects come into contact with each other, resulting in an exchange of energy and momentum.

What is the difference between an elastic and inelastic collision?

In an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved. In an inelastic collision, some of the kinetic energy is lost in the form of heat or sound.

How do you calculate the momentum of an object involved in a collision?

The momentum of an object is calculated by multiplying its mass by its velocity.

What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. This means that the total momentum of all objects involved in the collision remains constant.

How is the coefficient of restitution related to a collision?

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity after the collision to the relative velocity before the collision. A perfectly elastic collision has a coefficient of restitution of 1, while a completely inelastic collision has a coefficient of restitution of 0.

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