Calculating Block's Max Height from Impulse/Velocity

[joex444]

I found this to be the hardest question on my last test:

A 5kg block is resting on two tables such that the middle isn't resting on a part of a table. A 10g bullet is fired at 1000m/s from directly below the block perpendicular. The bullet exits the block going 400m/s. What was the maximum height of the block?

Here's what I did, the impulse was the change in momentum of the bullet, which would be .01kg(1000m/s)-.01(400m/s) = 6kgm/s. That impulse must have been imparted on the block, which would give it a final velocity of 1.2m/s (6kgm/s / 5kg). From there, y = [(v2^2 - v1^2)/-2a] letting a = -g = -9.8 and v2 = 0, v1 = 1.2, and I got 0.073m = 7.3cm.

Did I go about this right?

 

[Doc Al]

Looks good to me.

 

[quicknote]

Yes, your approach to solving this problem is correct. To find the maximum height of the block, you used the equation y = [(v2^2 - v1^2)/-2a], which is the formula for calculating the displacement of an object under constant acceleration. In this case, the acceleration is due to gravity with a value of -9.8m/s^2. The initial velocity of the block was 1.2m/s, and since it comes to a stop at the maximum height, the final velocity is 0m/s. Plugging in these values, you correctly calculated the maximum height to be 0.073m or 7.3cm. Great job!