Abullet of mass 0.017 kg traveling horizontally at a high speed of 290 m/s embeds itself in a block of mass 5 kg that is sitting at rest on a nearly frictionless surface.(a) what is the speed of the block after the bullet embeds itself in the block? (b) calculate the total translational kinetic energy before and after the collision.

(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

answer; i believe the correct answer is (40,000 j) good luck

it is called kinetic energy