Answer:
(a) vf = 0.98 m/s
(b) K₁ = 714.85 J : Total translational kinetic energy before the collision.
K₂= 2.41 J : Total translational kinetic energy after the collision.
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁ = 0.017 kg : mass of the bullet
m₂ = 5 kg : mass of the block
v₀₁ = 290 m/s : initial velocity of the bullet
v₀₂ = 0 : initial velocity of the block₂
(a) Speed of the block after the bullet embeds itself in the block
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = (m₁+ m₂)vf
vf: final velocity of the block
( 0.017)*( 290) + (5)*(0) = ( 0.017 + 5 )*vf
4.93+ 0 = ( 5.017 )*vf
vf = 4.93 / 5.017
vf = 0.98 m/s
b) Total translational kinetic energy before (K₁) and after the collision(K₂).
K₁ = 1/2(m₁*v₀₁² + m₂*v₀₂²)
K₁ = 1/2(0.017*(290)² + 5*(0)²) = 714.85 J
K₂= 1/2(m₁+ m₂)*vf²
K₂= 1/2(0.017+ 5)*(0.98)² = 2.41 J
