help physics please?

Question

When a bullet traveling at 739 m/s strikes a block of wood originally at rest on a frictionless surface, the bullet emerges from the other side of the block of wood traveling at 397 m/s. If the mass of the bullet is 4.79 g and the mass of the block is 715 g, what is the speed of the block after the collision?

Answer 1

momentum before = momentum after

momentum before: the bullet alone and the wood alone
momentum after: still the bullet alone and the wood alone

mv1 + mv2 = mv1 + mv2
(.00479)(739) + (.715)(0) = (.00479)(397) + (.715)v
3.53981 = 1.90163 + (.715)v
1.63818 = (.715)v
v = 2.29m/s

hope this helps.

Answer 2

This is about the conservation of (linear) momentum at the collision: if we write the momentum before the collision with p and after with p', then follows:
p = p'.
Before the collision, the block is at rest, so its momentum p2 = 0 and we only have the momentum of the bullet p1 = m1 x v1.
After the collision, both the bullet and the block move in the same direction, so we have:
p' = p1' + p'2 = m1 x v'1 + m2 x v'2.
With p = p' and after a little algebra, we have the speed of the block after the collision:
v'2 = (m1 x v1 ─ m1 x v'1) / m2 = m1 (v1 ─ v'1) / m2 =
= 4.79 g x (739 m/s ─ 397 m/s) / 715 g = 2.29 m/s.

Answer 3

This is a conservation of momentum equation. The momentum equation is Momentum=Mass * Velocity

I converted the mass to kg

(739*.00479)+(0*.715)=(397*.00479)+(Vblock * .715)
Velocity of the block = 2.29 m/s to the right