physics help?

Question

A 3.30-kg block of wood rests on the muzzle opening of a vertically oriented rifle, the stock of the rifle being firmly planted on the ground. When the rifle is fired, an 7.88-g bullet (velocity = 8.30 102 m/s, straight upward) becomes completely embedded in the block.
the velocity of the block/bullet system immediately after the collision.is 1.977

Ignoring air resistance, determine how high the block/bullet system rises above the muzzle opening of the rifle.
? m

Answer 1

First, momentum is conserved

7.88*830/1000=(3.33+7.88/1000)*1.977

6.54=6.56

That's close
the speed of the combined I get is 1.959 m/s

I will use 1.977 since that is what is asked


After the collision, use energy
m*g*h=.5*m*v^2
mass divides out and
h=.5*v^2/g
h=.5*1.977^2/9.81
0.199 m

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