A 1.60-g bullet, moving at 542 m/s, strikes a 0.249-kg piece of wood at rest on a frictionless table. The bullet sticks in the wood, and the combined mass moves slowly down the table.

Question

A 1.60-g bullet, moving at 542 m/s, strikes a 0.249-kg piece of wood at rest on a frictionless table. The bullet sticks in the wood, and the combined mass moves slowly down the table.
(a) Find the speed of the combination after the collision.
 m/s
(b) Find the kinetic energy of the bullet before the collision.
 J
(c) Find the kinetic energy of the combination after the collision.
 J
(d) How much kinetic energy did the bullet lose?
J
(e) What percent of the bullet's original kinetic energy is lost?
%

Elena · Answer

m1 = 0.0016 kg, v1 = 542 m/s, m2 =0.249, v2 = 0, u = ?

(a) m1•v1 +m2•v2 = (m1+m2) •u
u = m1•v1/(m1+m2).
(b) KE1 =m1•v1²/2.
(c) KE = ((m1+m2) •u²/2
(d) ΔKE = KE1 – KE.
(e) (ΔKE/KE1) •100%

Alica · Answer

I have never used these equations before, can you explain/ show me