# Two blocks collide

1. Mar 23, 2009

### Knfoster

1. The problem statement, all variables and given/known data

An 8.52g bullet is fired into a 385g block that is initially at rest at the edge of a table of h = 1.03 m height. The bullet remains in the block, and after the impact the block lands d = 2.15 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table? What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.

2. Relevant equations
m1v1=m2v2

3. The attempt at a solution

The velocity of the red card (the one moving) is 0 m/s, and the velocity of e blue cart just after collision is 3.13 m/s. But I don't know how to go about determining the initial speed of the bullet.

2. Mar 23, 2009

### LowlyPion

How long does it take for the block to drop 1.3m?

3. Mar 23, 2009

### sArGe99

Apply conservation of linear momentum. For finding out the time it takes to reach the ground, I would opt to go for a kinematic equation since the range is also given

4. Mar 23, 2009

### Knfoster

sry. I was copying and pasting... I meant to say that I've figured that the time it takes the block to reach the und is .458 s, and the initial horizontal velocity of the block is 4.69 m/s... What I don't know how to figure is the initial speed of the bullet...?

5. Mar 23, 2009

### LowlyPion

Well you have a perfectly inelastic collision.

You know the masses.
You know the final combined velocity.
You know the initial speed of the block at rest.
So ...

6. Mar 23, 2009

### Knfoster

so do I take the equation m1v1=(m1+m2)v2 ? And if so... what would my v2 be?

7. Mar 23, 2009

### sArGe99

Haven't you found that out?

8. Mar 23, 2009

### malta

I had a similar prob on mastering physics the other day,

Since

You know the masses.
You know the final combined velocity.
You know the initial speed of the block at rest.

then the total momentum= (mass block + mass bullet)final velocity

So the momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for total momentum , this time expressed as the total momentum of the system before the collision.

find this, set them equal to each other and solve for V

9. Mar 23, 2009

### Knfoster

THanks! I've figured it out. :)