Question

a bullet of mass m moving with a velocity u strike a block of mass M at rest and gets embedded in the block .the loss of kinetic energy in the impact is.

Answer 1

m = mass of bullet

u = initial velocity of bullet before the collision

M = mass of block

V = final velocity of bullet-block combination after collision

Using conservation of momentum

m u = (m + M) V

V = m u/(m + M)                                  eq-1

KE₁ = initial kinetic energy = (0.5) m u²

KE₂ = final kinetic energy = (0.5) (m + M) V²

Loss of kinetic energy is given as

ΔKE = KE₁ - KE₂

ΔKE = (0.5) m u²- (0.5) (m + M) V²

using eq-1

ΔKE = (0.5) m u²- (0.5) (m + M) (m u/(m + M) )²

ΔKE = (0.5) m u²- (0.5) m² u²/ (m + M)

ΔKE = ((0.5) m² u² + (0.5) m M u²-  (0.5) m² u²)/(m + M)

ΔKE = (0.5) m M u²/(m + M)

Answer 2

m1 = m

m2=M

u1=v

u2=0

let v1 = v2 = v3

total initial momentem of bullet and block = total final momentem

----------------------------------

m1×u1 m2 ×u2 = m1 ×v2+ m2 ×v2

mv = (M×0)=mv3 + Mv3

mv = ( m +M )v3

v3 =mv÷(m+m)