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

Answer 1

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

Answer 2

Solutions

Answer 3

kya ghatia explanation di hai. kuch bhi clear nahi hai. first baar kholi thi site ,bad experience bro. phir kabhi nahi open karu gi. 🙅🙅😠

Answer 4

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

Answer 5

Answer is wrong + explanation is super bad.....

Answer 6

Hugh

Answer 7

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