A 50 gram bullet moving with a velocity 10 m/s gets embedded into a 950 g stationary body . The loss in K.E. of the system will be
kya ghatia explanation di hai. kuch bhi clear nahi hai. first baar kholi thi site ,bad experience bro. phir kabhi nahi open karu gi. 🙅🙅ðŸ˜
Using conservation of momentum
mB*vB=mSystem*vSystem
vSystem=50*10/50+950
vSystem=0.5 (1/2)
KEloss={[0.5*mB*(vB)^2]-[0.5*mSys*(vSys)^2]}/[0.5*mB*(vB)^2]
mB=0.05kg,vB=10
mSys=1kg,vSys=0.5
Answer=0.95
Answer is wrong + explanation is super bad.....
Hugh
M1=50gms
=0.050kg
M2=950gms
.95kg
M1V1+m2v2=(m1+m2)v3
0.050x10+0=1v3
V3=0.5m/s
Loss in ke = final -initial
=1/2 x1x.5x.5-1/2x0.05x10x10
Solutions
initial momentum = 0.050 *10 = .5
final momentum = 1.00 * v = v
so v = 0.5 m/s
initial Ke = .5(.050)*100 = 2.5 Joules
final Ke = .5(1.00)*.25 = .125 Joules