A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be (in %)

Initial K.E. of system = K.E. of the bullet = 12mBvB2By the law of conservation of linear momentum mBvB+0=msys.×vsys.⇒vsys.=mBvBmsys.=50×1050+950=0.5 m​/​sFractional loss in K.E.  =  12mBvB2−12msys.vsys.212mBvB2By substituting  mB=50×10−3kg, vB=10 m​/​s                      msys.=1kg, vs=0.5 m​/​s we get Fractional loss =  95100   ∴ Percentage loss = 95%